With answers from 66 AI models.
กำหนดให้ p และ q เป็นประพจน์ ข้อใดต่อไปนี้ สามารถมีค่าความจริงได้มากกว่า 1 ค่า
Using dataset from OpenThaiGPT Evaluation Dataset.
ให้วิเคราะห์ทีละตัวเลือก:
a) (pvq) → q : มีค่าความจริงได้หลายค่า ขึ้นอยู่กับค่า p และ q b) (~p^q) → q : จะได้ค่าความจริงเป็น T เสมอ (Tautology) c) (~q^q) → p : จะได้ค่าความจริงเป็น T เสมอ เพราะ (~q^q) เป็น F เสมอ d) p v (p → q) : จะได้ค่าความจริงเป็น T เสมอ (Tautology) e) (pvq)(~p^~q) : มีค่าความจริงได้หลายค่า ขึ้นอยู่กับค่า p และ q
ตัวเลือกที่สามารถมีค่าความจริงได้มากกว่า 1 ค่า คือ ตัวเลือก a และ e
{"correct_answer_key":"a"}
Let me analyze each option to determine which one can have more than one truth value:
a) (pvq) → q: This is true when q is true, or when both p and q are false. It's false only when p is true and q is false.
b) (~p^q) → q: This is a tautology (always true) because the antecedent (~p^q) implies q.
c) (~q^q) → p: The antecedent (~q^q) is always false (contradiction), so this implication is always true regardless of p.
d) p v (p → q): This simplifies to p v (~p v q), which further simplifies to a tautology (always true).
e) (pvq) (~p^~q): This is a conjunction of (pvq) and (~p^~q). These two parts are contradictory - if either p or q is true, then (~p^~q) must be false. This expression is always false.
Looking at all options, only option a) (pvq) → q can have different truth values depending on the values of p and q.
{"correct_answer_key":"a"}
This question appears to be in Thai, asking about propositional logic. Let me translate first:
"Let p and q be propositions. Which of the following can have more than 1 truth value?"
Now, I'll analyze each option:
a) (pvq) → q This is "p OR q implies q" Let's make a truth table:
| p | q | p∨q | (p∨q)→q | |---|---|-----|---------| | T | T | T | T | | T | F | T | F | | F | T | T | T | | F | F | F | T |
So the formula (p∨q)→q can be true or false depending on the truth values of p and q. Specifically, it's false when p is true and q is false, and true otherwise. Therefore, this expression can have more than 1 truth value.
b) (~p^q) → q This is "NOT p AND q implies q" Truth table:
| p | q | ~p | ~p∧q | (~p∧q)→q | |---|---|----|----- |----------| | T | T | F | F | T | | T | F | F | F | T | | F | T | T | T | T | | F | F | T | F | T |
In all cases, the expression is true. Therefore, this expression has only 1 truth value (always true).
c) (~q^q) → p This is "NOT q AND q implies p"
| p | q | ~q | ~q∧q | (~q∧q)→p | |---|---|----|------|----------| | T | T | F | F | T | | T | F | T | F | T | | F | T | F | F | T | | F | F | T | F | T |
Note that ~q∧q is always false (contradiction), and false implies anything is true. So this is always true regardless of p and q. Therefore, it has only 1 truth value.
d) p v (p → q) This is "p OR (p implies q)"
| p | q | p→q | p∨(p→q) | |---|---|-----|---------| | T | T | T | T | | T | F | F | T | | F | T | T | T | | F | F | T | T |
The expression is always true. Therefore, it has only 1 truth value.
e) (pvq) (~p^~q) I think there might be some operation missing here. Let's assume it's a conjunction: (pvq) ∧ (~p^~q)
| p | q | p∨q | ~p | ~q | ~p∧~q | (p∨q)∧(~p∧~q) | |---|---|-----|----|----|-------|---------------| | T | T | T | F | F | F | F | | T | F | T | F | T | F | F | | F | T | T | T | F | F | F | | F | F | F | T | T | T | F |
Note: (p∨q) and (~p∧~q) are contradictory - if one is true, the other must be false. So their conjunction is always false.
However, I notice in my work that (p∨q)∧(~p∧~q) is actually always false. Let me double-check this.
We have:
These can't be true at the same time. So (p∨q)∧(~p∧~q) is always false.
If the question meant some other operation between (pvq) and (~p^~q), I would need more information.
Looking at my analysis: a) (pvq) → q: Can be T or F b) (~p^q) → q: Always T c) (~q^q) → p: Always T d) p v (p → q): Always T e) (pvq) (~p^~q): If conjunction, always F
The question asks which can have more than 1 truth value. From my analysis, only option a) (pvq) → q can have more than one truth value (it can be T or F).
Therefore, the answer is a.
Wait, I think I might have made a mistake in option e. Let me double-check it once more.
For option e) (pvq) (~p^~q), I assumed a conjunction between (pvq) and (~p^~q). If that's the case, since these two are contradictory, their conjunction is always false.
However, if option e means something different, like a material conditional or some other operation, I would need to re-evaluate.
Actually, I'm not entirely sure what the operation between (pvq) and (~p^~q) is in option e. It might be a material implication (→), a biconditional (↔), or even something else. Without knowing the operation, I can't determine the truth values.
Let me reconsider option e assuming various operations:
If it's material implication (pvq) → (~p^~q):
| p | q | p∨q | ~p | ~q | ~p∧~q | (p∨q)→(~p∧~q) | |---|---|-----|----|----|-------|---------------| | T | T | T | F | F | F | F | | T | F | T | F | T | F | F | | F | T | T | T | F | F | F | | F | F | F | T | T | T | T |
So if it's a material implication, it can be T or F, having more than 1 truth value.
If it's a biconditional (pvq) ↔ (~p^~q):
| p | q | p∨q | ~p | ~q | ~p∧~q | (p∨q)↔(~p∧~q) | |---|---|-----|----|----|-------|---------------| | T | T | T | F | F | F | F | | T | F | T | F | T | F | F | | F | T | T | T | F | F | F | | F | F | F | T | T | T | T |
For a biconditional, it can also be T or F, having more than 1 truth value.
Given the ambiguity in option e, and based on the analysis of other options, options a) and potentially e) can have more than 1 truth value.
But since we have to pick one, I'll go with option a) (pvq) → q.
Actually, let me be more systematic to ensure I'm not missing anything. I'll go through each option one more time:
a) (pvq) → q
| p | q | p∨q | (p∨q)→q | |---|---|-----|---------| | T | T | T | T | | T | F | T | F | | F | T | T | T | | F | F | F | T |
It can be T or F, so it has more than 1 truth value.
b) (~p^q) → q
| p | q | ~p | ~p∧q | (~p∧q)→q | |---|---|----|----- |----------| | T | T | F | F | T | | T | F | F | F | T | | F | T | T | T | T | | F | F | T | F | T |
It's always T, so it has only 1 truth value.
c) (~q^q) → p
| p | q | ~q | ~q∧q | (~q∧q)→p | |---|---|----|------|----------| | T | T | F | F | T | | T | F | T | F | T | | F | T | F | F | T | | F | F | T | F | T |
It's always T, so it has only 1 truth value.
d) p v (p → q)
| p | q | p→q | p∨(p→q) | |---|---|-----|---------| | T | T | T | T | | T | F | F | T | | F | T | T | T | | F | F | T | T |
It's always T, so it has only 1 truth value.
e) (pvq) (~p^~q)
Without a clear operation, it's hard to determine. But let's assume it's a conjunction:
| p | q | p∨q | ~p | ~q | ~p∧~q | (p∨q)∧(~p∧~q) | |---|---|-----|----|----|-------|---------------| | T | T | T | F | F | F | F | | T | F | T | F | T | F | F | | F | T | T | T | F | F | F | | F | F | F | T | T | T | F |
Wait, I think I made an error in the last row. Let's correct it:
| p | q | p∨q | ~p | ~q | ~p∧~q | (p∨q)∧(~p∧~q) | |---|---|-----|----|----|-------|---------------| | T | T | T | F | F | F | F | | T | F | T | F | T | F | F | | F | T | T | T | F | F | F | | F | F | F | T | T | T | F |
Actually, in the last row (p=F, q=F), both p∨q and ~p∧~q are true? No, that can't be right. Let me double-check:
For the last row where p=F and q=F:
So, the result for (p∨q)∧(~p∧~q) would be F∧T = F.
Ok, so the corrected table is:
| p | q | p∨q | ~p | ~q | ~p∧~q | (p∨q)∧(~p∧~q) | |---|---|-----|----|----|-------|---------------| | T | T | T | F | F | F | F | | T | F | T | F | T | F | F | | F | T | T | T | F | F | F | | F | F | F | T | T | T | F |
Actually, I think I need to revisit the last row for (p∨q)∧(~p∧~q). For p=F, q=F:
Hmm, I still got F for the last row. Let me analyze more carefully:
For p=F, q=F:
So, for option e, if it's a conjunction, it's always F, having only 1 truth value.
Actually, I think there might be an operation missing in option e. Let's reconsider the original:
e) (pvq) (~p^~q)
If we assume it's a multiplication or a logical AND, it would be (pvq) AND (~p^~q).
I've already analyzed this and found that it's always F, having only 1 truth value.
But, if we consider it as a product (like in boolean algebra), we'd get terms like: p∧~p∧~q, q∧~p∧~q. Both of these are contradictory, and so, the entire expression would always be F.
So, for option e, it seems to have only 1 truth value.
I've now analyzed all options: a) (pvq) → q: Can be T or F b) (~p^q) → q: Always T c) (~q^q) → p: Always T d) p v (p → q): Always T e) (pvq) (~p^~q): Always F (assuming conjunction or boolean product)
The question asks which can have more than 1 truth value. From my analysis, only option a) (pvq) → q can have more than one truth value (it can be T or F).
Therefore, the answer is a.
Actually, I want to double-check my analysis of option e. Let me restate it:
e) (pvq) (~p^~q)
Without an explicit operator, it's hard to determine the exact meaning. In propositional logic, we usually have operations like AND, OR, IMPLIES, etc.
But I'll assume a few possible interpretations:
If it's a conjunction (AND): (pvq) ∧ (~p^~q) I've already analyzed this and found it's always F.
If it's a material implication (IMPLIES): (pvq) → (~p^~q)
| p | q | p∨q | ~p | ~q | ~p∧~q | (p∨q)→(~p∧~q) | |---|---|-----|----|----|-------|---------------| | T | T | T | F | F | F | F | | T | F | T | F | T | F | F | | F | T | T | T | F | F | F | | F | F | F | T | T | T | T |
This can be T or F, having more than 1 truth value.
If it's a biconditional (IFF): (pvq) ↔ (~p^~q)
| p | q | p∨q | ~p | ~q | ~p∧~q | (p∨q)↔(~p∧~q) | |---|---|-----|----|----|-------|---------------| | T | T | T | F | F | F | F | | T | F | T | F | T | F | F | | F | T | T | T | F | F | F | | F | F | F | T | T | T | T |
This can also be T or F, having more than 1 truth value.
I had overlooked these other interpretations. However, without a clear operator, it's hard to be sure which is the correct interpretation for option e.
Given the information at hand and considering all possible interpretations for option e, both options a) and e) can potentially have more than 1 truth value.
Actually, I think the most likely interpretation for e) (pvq) (~p^~q) is a conjunction, which would make it (pvq) ∧ (~p^~q). In this case, as I've shown, it's always F, having only 1 truth value.
Let me also check option e) with another possible interpretation: what if it's a disjunction (OR)?
(pvq) ∨ (~p^~q)
| p | q | p∨q | ~p | ~q | ~p∧~q | (p∨q)∨(~p∧~q) | |---|---|-----|----|----|-------|---------------| | T | T | T | F | F | F | T | | T | F | T | F | T | F | T | | F | T | T | T | F | F | T | | F | F | F | T | T | T | T |
If it's a disjunction, the expression is always T, having only 1 truth value.
Based on my analysis, of all the listed options, only a) (pvq) → q can have more than 1 truth value.
Therefore, the answer is a.
Actually, reviewing how Thai often represents and translates logical operators, I notice something about the ambiguous structure of option e. It's possible that the intended notation means disjunction of the distributed form (pvq)∧(~p)∧(~q). Let me check this:
(p∨q)∧(~p)∧(~q)
| p | q | p∨q | ~p | ~q | (p∨q)∧(~p)∧(~q) | |---|---|-----|----|----|-----------------| | T | T | T | F | F | F | | T | F | T | F | T | F | | F | T | T | T | F | F | | F | F | F | T | T | F |
But in the last row (p=F, q=F), I think there's an issue:
All the cases yield F, so this interpretation also has only 1 truth value.
I think I'm overcomplicating this. Given the original question and the typical notation in propositional logic, option a) (pvq) → q is the most straightforward choice that can have more than 1 truth value.
Therefore, the answer is a.
Hmm, actually, let me check one more thing with option e.
Option e: (pvq) (~p^~q)
Given that there's no explicit operator between (pvq) and (~p^~q), it's unusual notation. However, the only way I can make sense of this in standard propositional logic is to interpret it as the conjunction of the two expressions:
(pvq) ∧ (~p^~q)
I've already analyzed this and found it to be always F.
But what if it means (p∨q)∧(~p)∧(~q)? Let's check:
| p | q | p∨q | ~p | ~q | (p∨q)∧(~p)∧(~q) | |---|---|-----|----|----|-----------------| | T | T | T | F | F | F | | T | F | T | F | T | F | | F | T | T | T | F | F | | F | F | F | T | T | F |
Ah, I think I've been too perfectionist about this. Option e is most likely intended as (pvq) ∧ (~p^~q), which is always F. So, option e has only 1 truth value.
Therefore, option a) is the correct answer to the question, as it's the only one among the given options that can have more than 1 truth value.
Wait, let's double check option e one more time with a slightly different approach, since it's ambiguous.
Option e: (pvq) (~p^~q)
If this is intended as the conjunction of the two expressions without using an explicit ∧ symbol, it would mean:
(p∨q) ∧ (~p∧~q)
Let's check its truth value:
Can both (p∨q) and (~p∧~q) be true simultaneously? No, they're logically contradictory. If both p and q are false, then (p∨q) is false. And if at least one of p or q is true, then (~p∧~q) is false.
So, this expression is always false, having only 1 truth value.
My final conclusion: option a) (pvq) → q is the only option among the given ones that can have more than 1 truth value.
Therefore, the answer is a.
I need to determine which propositional logic formula can have more than one truth value. Let me analyze each option:
a) (pvq) → q This means "p OR q implies q" When p is T and q is F: (TvF) → F = T → F = F When p is F and q is T: (FvT) → T = T → T = T So this can be both T or F depending on the values of p and q.
b) (~p^q) → q This is a tautology (always true) because if q is true, the implication is true, and if q is false, the premise ~p^q is false, making the implication true.
c) (~q^q) → p This is always true because ~q^q is always false (contradiction), and false implies anything is true.
d) p v (p → q) This is always true regardless of p and q's values.
e) (pvq) (~p^~q) If interpreted as conjunction, this is always false as p∨q and ~p∧~q cannot be true simultaneously.
Only option a can have more than one truth value.
{"correct_answer_key":"a"}
Let me analyze each logical expression to determine which one can have more than one truth value (i.e., which is not a tautology or contradiction).
For each option, I'll check if it's always true, always false, or can be both depending on the values of p and q.
a) (p∨q) → q: This can be false when p is true and q is false, but true in other cases. So it can have different truth values.
b) (~p∧q) → q: When (~p∧q) is true, q must be true, making the implication true. When (~p∧q) is false, the implication is also true. This is a tautology (always true).
c) (~q∧q) → p: The antecedent (~q∧q) is always false (contradiction), making this implication always true regardless of p. This is a tautology.
d) p ∨ (p → q): Using logical equivalence, p → q ≡ ~p ∨ q, so this becomes p ∨ (~p ∨ q) ≡ (p ∨ ~p) ∨ q ≡ T ∨ q ≡ T. This is a tautology.
e) (p∨q) ∧ (~p∧~q): This states that (p or q) and (not p and not q) are both true, which is impossible. This is a contradiction (always false).
Option a) is the only expression that can have different truth values depending on the values of p and q.
{"correct_answer_key": "a"}
Let me analyze each logical expression to determine which one can have more than one truth value (i.e., which one is not a tautology or contradiction).
Let me create truth tables for each option:
For p and q, we have 4 possible combinations:
Option a: (p∨q) → q
Option b: (~p∧q) → q
Option c: (~q∧q) → p Note that (~q∧q) is always false (contradiction)
When p=T, q=T: (~T∧T) → T = (F∧T) → T = F → T = T
When p=T, q=F: (~F∧F) → T = (T∧F) → T = F → T = T
When p=F, q=T: (~T∧T) → F = (F∧T) → F = F → F = T
When p=F, q=F: (~F∧F) → F = (T∧F) → F = F → F = T Result: T, T, T, T (always true - tautology)
Option d: p ∨ (p → q)
I can conclude that only Option a has varying truth values, making it the most interesting logical expression. I'll complete the partial thought from the previous thinking first:
Analyzing the truth values, I see each option has a distinct logical behavior. Some remain constant, while others demonstrate variability. The systematic evaluation exposes the nuanced logical relationships between different propositional combinations.
The critical insight emerges from option a's unique characteristic of potential truth value variation, distinguishing it from the other uniformly true or false options. This variability makes it the most intriguing and distinctive choice in the logical analysis.
Looking at each logical expression, I need to determine which one is neither a tautology (always true) nor a contradiction (always false).
Let me evaluate each option using truth tables:
Option a: (p∨q) → q
Option b: (~p∧q) → q This is always true (tautology) because whenever (~p∧q) is true, q must be true.
Option c: (~q∧q) → p (~q∧q) is always false (contradiction), so this implication is always true (tautology).
Option d: p ∨ (p → q) This is always true (tautology) - when p is true, the disjunction is true; when p is false, (p → q) is true.
Option e: (p∨q) ∧ (~p∧~q) This is always false (contradiction) as (p∨q) and (~p∧~q) are contradictory.
Only option a can have different truth values depending on the values of p and q.
{"correct_answer_key": "a"}
ให้ฉันวิเคราะห์แต่ละตัวเลือกโดยใช้ตารางค่าความจริง:
a) (p∨q) → q: เมื่อ p∨q เป็นจริง แต่ q เป็นเท็จ (p=T, q=F) จะได้ F; กรณีอื่นๆ ได้ T b) (~p∧q) → q: เมื่อ ~p∧q เป็นจริง q ต้องเป็นจริง ดังนั้นจะได้ T เสมอ c) (~q∧q) → p: ~q∧q เป็นเท็จเสมอ (contradiction) ดังนั้น F→p ได้ T เสมอ d) p ∨ (p → q): เมื่อ p=T จะได้ T; เมื่อ p=F, (F→q) ได้ T เสมอ ดังนั้นได้ T เสมอ e) (p∨q) ∧ (~p∧~q): นี่คือ (p∨q) ∧ ~(p∨q) ซึ่งเป็น contradiction ได้ F เสมอ
ตัวเลือก a เท่านั้นที่สามารถมีค่าความจริงได้ทั้ง T และ F
{"correct_answer_key": "a"}
This is a logic question in Thai. Let me translate and understand it first.
The question asks: "Given that p and q are propositions, which of the following can have more than 1 truth value?"
Let me analyze each option by creating truth tables:
a) (p∨q) → q Truth table: p | q | p∨q | (p∨q)→q T | T | T | T T | F | T | F F | T | T | T F | F | F | T
This can have both T and F values.
b) (~p∧q) → q Truth table: p | q | ~p | ~p∧q | (~p∧q)→q T | T | F | F | T T | F | F | F | T F | T | T | T | T F | F | T | F | T
This is always true (tautology).
c) (~q∧q) → p Note that ~q∧q is always false (contradiction) Truth table: p | q | ~q | ~q∧q | (~q∧q)→p T | T | F | F | T T | F | T | F | T F | T | F | F | T F | F | T | F | T
This is always true (tautology) because a false antecedent makes the implication true.
d) p ∨ (p → q) Truth table: p | q | p→q | p∨(p→q) T | T | T | T T | F | F | T F | T | T | T F | F | T | T
This is always true (tautology).
e) (p∨q) ∧ (~p∧~q)
I'll complete the truth table to analyze this proposition carefully. The last row is incomplete, so I'll fill it in and examine the logical possibilities systematically.
The truth table reveals this statement is fundamentally a contradiction. When I evaluate each option, I notice only option a allows multiple truth values, while the others are consistently true or false.
This means the statement has no scenarios where it can simultaneously be true and false, except for option a. The logical structure ensures a definitive outcome for most options, making option a uniquely flexible in its truth assignment.
ให้ฉันวิเคราะห์แต่ละตัวเลือกด้วยตารางค่าความจริง:
สำหรับประพจน์ที่สามารถมีค่าความจริงได้มากกว่า 1 ค่า หมายถึงไม่ใช่ tautology (จริงเสมอ) หรือ contradiction (เท็จเสมอ)
a) (p∨q) → q: เมื่อ p=T, q=F จะได้ค่าเป็นเท็จ แต่เมื่อ p=F, q=T จะได้ค่าเป็นจริง (มีได้ทั้ง T และ F)
b) (~p∧q) → q: เป็นจริงเสมอ (tautology) เพราะเมื่อ ~p∧q เป็นจริง q ต้องเป็นจริง
c) (~q∧q) → p: เป็นจริงเสมอ (tautology) เพราะ ~q∧q เป็นเท็จเสมอ ทำให้ประพจน์เงื่อนไขเป็นจริงเสมอ
d) p ∨ (p → q): เป็นจริงเสมอ (tautology) เมื่อ p เป็นเท็จ (p → q) จะเป็นจริง
e) (p∨q) ∧ (~p∧~q): เป็นเท็จเสมอ (contradiction) เพราะเป็นการ AND ของประพจน์ที่ขัดแย้งกัน
{"correct_answer_key":"a"}
ให้ฉันวิเคราะห์แต่ละตัวเลือกโดยการสร้างตารางค่าความจริง:
สำหรับประพจน์ p และ q จะมีค่าความจริงได้ 4 แบบ: (T,T), (T,F), (F,T), (F,F)
a) (p∨q) → q: เมื่อ p=T, q=F จะได้ (T∨F) → F = T → F = F, แต่กรณีอื่นเป็น T b) (~p∧q) → q: เมื่อ q=T แล้ว (~p∧q) → q จะเป็น T เสมอ, เมื่อ q=F แล้ว (~p∧q) เป็น F ทำให้ F → F = T c) (~q∧q) → p: เนื่องจาก (~q∧q) เป็น contradiction (เป็น F เสมอ) ดังนั้น F → p จะเป็น T เสมอไม่ว่า p จะเป็นอะไร d) p ∨ (p → q): เมื่อ p=T จะได้ T ∨ (อะไรก็ได้) = T, เมื่อ p=F จะได้ F ∨ (F → q) = F ∨ T = T e) (p∨q) ∧ (~p∧~q): นี่คือ contradiction เพราะ (p∨q) และ (~p∧~q) ไม่สามารถเป็นจริงพร้อมกันได้
เฉพาะตัวเลือก a เท่านั้นที่มีค่าความจริงได้ทั้ง T และ F ขึ้นอยู่กับค่าของ p และ q
{"correct_answer_key":"a"}
This is a logic question in Thai. Let me translate it first:
"Given that p and q are propositions. Which of the following can have more than 1 truth value?"
Let me analyze each option by checking their truth tables:
a) (p∨q) → q Let me check all combinations:
So this can be T or F, meaning it can have more than 1 truth value.
b) (~p∧q) → q Let me check:
This is always T (tautology), so it has only 1 truth value.
c) (~q∧q) → p First, ~q∧q is always F (contradiction), so F → p is always T regardless of p. This is always T (tautology), so it has only 1 truth value.
d) p ∨ (p → q) Let me check:
p=T, q=T: T ∨ (T → T) = T ∨ T = T
p=T, q=F: T ∨ (T → F) = T ∨ F = T
p=F, q=T: F ∨ (F → T) = F ∨ T = T
p=F, q=F: F ∨ (F → F) = F ∨ T = T
This is always T (tautology), so it has only 1 truth value.
e) (p∨q) ∧ (~p∧~q) I'll continue analyzing this option's truth values carefully. The expression involves combining disjunction and conjunction with complemented propositions, which suggests a complex logical interaction.
When checking each possible combination of truth values, I notice a consistent pattern. For every scenario - whether p and q are both true, both false, or mixed - the result is always F. This means the statement is a contradiction, producing only one truth value across all possible inputs.
The logical structure ensures that the disjunction and negation cancel out any potential true outcomes, leading to a consistently false result. This makes the expression fundamentally self-contradictory, regardless of the initial truth values of p and q.
a) (p∨q) → q I'll examine this implication carefully. The truth table reveals an interesting characteristic: this proposition doesn't maintain a single truth value. Some scenarios yield true results, while others produce false outcomes.
Specifically, when p is false and q is false, the implication surprisingly evaluates to true. This variability means the statement isn't a tautology or a contradiction, but can take on multiple truth values depending on the initial conditions.
The logical structure shows that the truth of the disjunction combined with the subsequent implication creates a nuanced logical relationship that doesn't conform to a single, predictable truth value.
Looking at this logic question in Thai, I need to determine which proposition can have more than one truth value (i.e., can be either true or false depending on the truth values of p and q).
Let me analyze each option by checking truth tables:
a) (p∨q) → q: When p=T, q=F, we get (T∨F) → F = T → F = F. When p=F, q=T, we get (F∨T) → T = T → T = T. This can be both true and false.
b) (~p∧q) → q: This is always true (tautology) because if the antecedent (~p∧q) is true, then q must be true, making the consequent true.
c) (~q∧q) → p: Since (~q∧q) is always false (contradiction), this becomes F → p, which is always true.
d) p ∨ (p → q): This is always true (tautology) because either p is true (making the disjunction true) or p is false (making p → q true).
e) (p∨q) ∧ (~p∧~q): This is always false (contradiction) because it requires both (p or q) and (neither p nor q) to be true simultaneously.
Only option (a) can have different truth values depending on the values of p and q.
{"correct_answer_key":"a"}
To determine which statement can have more than one truth value, we need to analyze each option based on the truth tables of logical connectives.
Option A: (pvq) → q
Option B: (~p^q) → q
Option C: (~q^q) → p
Option D: p v (p → q)
Option E: (pvq) → (~p^~q)
Correct Answer:
{"correct_answer_key":"d"}
This question is about propositional logic and asks which of the given logical expressions can have more than one possible truth value (i.e., is not a tautology or a contradiction, but depends on the truth values of ( p ) and ( q )).
We need to analyze each option to see if its truth value can vary (i.e., it is not always true or not always false).
Option A: ((p \lor q) \rightarrow q)
Option B: ((\neg p \land q) \rightarrow q)
Option C: ((\neg q \land q) \rightarrow p)
Option D: ( p \lor (p \rightarrow q) )
Option E: ((p \lor q) \land (\neg p \land \neg q))
Only Option A can have more than one truth value (sometimes true, sometimes false).
{"correct_answer_key":"a"}
Explanation:
The question asks which of the given logical expressions can have more than one truth value (i.e., it is not a tautology and not a contradiction, so its truth value depends on the truth values of p and q). We need to check each option to see if it is contingent (can be true or false).
Let p and q be propositions (can be true or false). We analyze each option:
a) (p ∨ q) → q
This is equivalent to ¬(p ∨ q) ∨ q ≡ (¬p ∧ ¬q) ∨ q.
If q is true, it is true. If q is false, it becomes (¬p ∧ true) ∨ false ≡ ¬p. So it can be true (if p false) or false (if p true). Thus, it has more than one truth value.
b) (¬p ∧ q) → q
This is equivalent to ¬(¬p ∧ q) ∨ q ≡ (p ∨ ¬q) ∨ q ≡ p ∨ (¬q ∨ q) ≡ p ∨ true ≡ true.
This is a tautology (always true). So it has only one truth value.
c) (¬q ∧ q) → p
¬q ∧ q is always false (contradiction). So (false → p) is always true (since false implies anything). This is a tautology. Only one truth value.
d) p ∨ (p → q)
p → q is equivalent to ¬p ∨ q. So this becomes p ∨ (¬p ∨ q) ≡ (p ∨ ¬p) ∨ q ≡ true ∨ q ≡ true.
Tautology. Only one truth value.
e) (p ∨ q) ∧ (¬p ∧ ¬q) [Note: The original says "(pvq) (~p^~q)" which likely means conjunction, so (p ∨ q) ∧ (¬p ∧ ¬q)]
This is equivalent to (p ∨ q) ∧ ¬(p ∨ q) (by De Morgan) which is always false (contradiction). So only one truth value.
Therefore, only option a) can have more than one truth value (it is contingent).
{"correct_answer_key": "a"}
The question asks which logical expression can have more than one truth value (i.e., it is not a tautology or contradiction) for propositions (p) and (q).
Option a: ((p \lor q) \to q)
This can be true or false depending on (p) and (q). For example:
Option b: ((\neg p \land q) \to q)
This is always true (tautology). For example:
Option c: ((\neg q \land q) \to p)
(\neg q \land q) is always false (contradiction), and (\text{false} \to p) is always true (tautology). For example:
Option d: (p \lor (p \to q))
(p \to q) is equivalent to (\neg p \lor q), so (p \lor (\neg p \lor q) = (p \lor \neg p) \lor q = \text{true} \lor q = \text{true}) (tautology). Thus, it has only one truth value.
Option e: ((p \lor q) \land (\neg p \land \neg q))
(\neg p \land \neg q) is equivalent to (\neg (p \lor q)), so this is ((p \lor q) \land \neg (p \lor q)), which is always false (contradiction). For example:
Only option a can have more than one truth value.
{"correct_answer_key": "a"}
First, the question is in Thai. It says: "Let p and q be propositions. Which of the following can have more than one truth value?" I need to evaluate each option to see which one can be true or false depending on the truth values of p and q.
The options are:
a) (p ∨ q) → q
b) (~p ∧ q) → q
c) (~q ∧ q) → p
d) p ∨ (p → q)
e) (p ∨ q) (~p ∧ ~q) // This might be missing an operator; probably it's (p ∨ q) ∧ (~p ∧ ~q) or something similar. Looking back, it says "(pvq) (~p^~q)" which might be a typo. In logic, when two expressions are next to each other, it often implies conjunction, so I'll assume it's (p ∨ q) ∧ (~p ∧ ~q).
Now, I need to check for each expression whether its truth value can vary based on p and q. That means it shouldn't be a tautology (always true) or a contradiction (always false); it should be contingent.
Let me analyze each one step by step.
Start with option a: (p ∨ q) → q
I'll make a truth table for this.
| p | q | p ∨ q | (p ∨ q) → q |
| T | T | T | T |
| T | F | T | F |
| F | T | T | T |
| F | F | F | T |
So, the truth values are: T, F, T, T. It can be true or false. Specifically, when p is true and q is false, it's false; otherwise true. So, it has more than one truth value. But I need to check all options; maybe others are always true or false.
Now option b: (~p ∧ q) → q
Truth table:
| p | q | ~p | ~p ∧ q | (~p ∧ q) → q |
| T | T | F | F | T |
| T | F | F | F | T |
| F | T | T | T | T | // because if antecedent is true, q is true, so implication true
| F | F | T | F | T |
So, it's always true. Tautology. So, it has only one truth value: always true.
Option c: (~q ∧ q) → p
~q ∧ q is always false because it's a contradiction. So, false → p. Since false implies anything is true, this is always true. Tautology.
Truth table:
| p | q | ~q | ~q ∧ q | (~q ∧ q) → p |
| T | T | F | F | T |
| T | F | T | F | T |
| F | T | F | F | T |
| F | F | T | F | T |
Always true. So, only one truth value.
Option d: p ∨ (p → q)
p → q is equivalent to ~p ∨ q.
So, p ∨ (~p ∨ q) which is (p ∨ ~p) ∨ q, which is true ∨ q, which is always true. Tautology.
Truth table:
| p | q | p → q | p ∨ (p → q) |
| T | T | T | T |
| T | F | F | T | // because p is true, so true ∨ false is true
| F | T | T | T |
| F | F | T | T | // p is false, but p → q is true (since false → false is true), so true ∨ false is true
Always true. So, tautology.
Now option e: (p ∨ q) (~p ∧ ~q)
As I thought, it's probably (p ∨ q) ∧ (~p ∧ ~q). But let's confirm the notation. It says "(pvq) (~p^~q)" which might be missing an operator. In some contexts, it could be implication, but without an operator, it's likely conjunction. I'll assume it's (p ∨ q) ∧ (~p ∧ ~q).
Now, ~p ∧ ~q is the negation of p ∨ q, by De Morgan's law. So, (p ∨ q) ∧ ¬(p ∨ q) which is always false. Contradiction.
Truth table:
| p | q | p ∨ q | ~p | ~q | ~p ∧ ~q | (p ∨ q) ∧ (~p ∧ ~q) |
| T | T | T | F | F | F | F |
| T | F | T | F | T | F | F |
| F | T | T | T | F | F | F |
| F | F | F | T | T | T | F | // because p ∨ q is false, so false ∧ true is false
So, always false. Contradiction. Only one truth value.
But option a can have more than one truth value. It can be true or false.
Now, looking back at option e, the notation might be ambiguous. It says "(pvq) (~p^~q)" which could be interpreted as (p ∨ q) → (~p ∧ ~q) or something else. But in the original, it's written without an operator, which is odd. In the JSON, it's "(pvq) (~p^~q)" so probably it's meant to be a single expression, but in logic, we need an operator.
Perhaps it's (p ∨ q) ∧ (~p ∧ ~q), as I thought, which is always false.
Maybe it's (p ∨ q) ≡ (~p ∧ ~q) or something, but that's not written.
Another possibility: in some notations, juxtaposition might imply conjunction, so I think my assumption is correct.
But let's double-check the question: "e":"(pvq) (~p^~q)"
It might be a typo, and it should have an operator like → or ∧.
But based on standard interpretation, it's likely conjunction.
Perhaps it's meant to be (p ∨ q) → (~p ∧ ~q), but that would be different.
Let me check that possibility.
If e is (p ∨ q) → (~p ∧ ~q)
Truth table:
| p | q | p ∨ q | ~p | ~q | ~p ∧ ~q | (p ∨ q) → (~p ∧ ~q) |
| T | T | T | F | F | F | F |
| T | F | T | F | T | F | F |
| F | T | T | T | F | F | F |
| F | F | F | T | T | T | T |
So, it can be false or true. When p and q both false, it's true; otherwise false. So, it has more than one truth value.
But in the option, it's written without an operator, so it's ambiguous.
Now, looking back at the original JSON: "e":"(pvq) (~p^~q)"
In logic problems, sometimes they might mean equivalence or something, but here it's not specified.
Perhaps it's a mistake, and it should be with an operator.
But let's see the other options: all a,b,c,d have operators like → or v or ^, so e might be missing one.
In option e, it has "(pvq)" and then "(~p^~q)" without a connective. That might be a typo.
Perhaps it's (p ∨ q) ∧ (~p ∧ ~q), which is always false.
But in that case, only option a has varying truth values.
Now, for option d, I thought it was tautology, but let me confirm.
p ∨ (p → q)
p → q is false only when p true and q false, so in that case, p is true, so p ∨ false is true. Otherwise, always true. Yes.
But option a is not always true; it can be false.
Now, for option c, always true.
Option b, always true.
So, only option a can have more than one truth value.
But let's confirm option e. If it is (p ∨ q) → (~p ∧ ~q), then it can be true or false, so it might also have more than one value.
But the notation doesn't have an arrow; it's just written next to each other.
In the JSON, it's given as "(pvq) (~p^~q)" which might be intended to be read as (p ∨ q) and (~p ∧ ~q) with an implied operator, but which one?
In multiple choice, sometimes they might have a space meaning conjunction, but that would be always false.
Perhaps it's a disjunction or something.
Another thought: in some notations, "pvq" might be written, but here it's separate.
Look at the input: "e":"(pvq) (~p^~q)"
Notice that in other options, they use "^" for conjunction, so in e, "(~p^~q)" is clearly conjunction.
But between "(pvq)" and "(~p^~q)", there is no operator.
This is likely a typo in the question. Perhaps it's meant to be (p ∨ q) ↔ (~p ∧ ~q) or something.
But let's see the context. The question is about truth values, and option a already works.
Perhaps for e, it is (p ∨ q) ∧ (~p ∧ ~q), which is always false, so only one truth value.
But to be thorough, let's assume common mistakes.
Perhaps the space implies a connective like implication, but that's not standard.
In the original Thai, it might be clear, but I don't read Thai.
Another idea: in option e, it might be (p ∨ q) ⊕ (~p ∧ ~q) or something, but not specified.
Given that, and since option a clearly has multiple truth values, and others seem tautologies or contradictions, probably a is the answer.
But let's list all:
a: contingent (can be T or F)
b: tautology (always T)
c: tautology (always T)
d: tautology (always T)
e: if conjunction, always F; if implication, contingent.
But in the notation, it's not written with an operator, so likely it's a mistake, and we should go with a.
Perhaps in the context, " can have more than one truth value" means that for some assignments it's true, for some false, so a and e (if implication) both.
But e is not properly written.
Look back at the JSON: "e":"(pvq) (~p^~q)"
Notice that in the question, it's written with spaces, but in logic, we need an operator.
Perhaps it is meant to be (p ∨ q) ∧ (~p ∧ ~q), which is always false.
To confirm, let's see if there is any option that is not tautology or contradiction.
Option a is the only one that is contingent.
For option d, p ∨ (p → q) is indeed always true, as I said.
p → q is equivalent to ~p ∨ q, so p ∨ (~p ∨ q) = (p ∨ ~p) ∨ q = true ∨ q = true.
Yes.
So, only a can have different truth values.
Therefore, the answer should be a.
But let's see the choices: a, b, c, d, e.
So, for e, if we interpret it as is, it might be invalid, but probably it's always false.
Thus, I think a is correct.
So, the correct answer key is "a".
The question asks which logical expression can have more than one truth value (i.e., it is contingent, not a tautology or contradiction).
Thus, only option a can have more than one truth value.
{"correct_answer_key": "a"}
The question asks which of the following propositions can have more than one truth value (meaning it can be either true or false, depending on the truth values of p and q). Let's analyze each option:
a) (p∨q) → q: If q is true, the implication is true. If q is false, the implication is true only if p is also false. So this can be true or false.
b) (~p∧q) → q: If q is true, the implication is true. If q is false, the antecedent (~p∧q) is false, making the implication true. This is always true (a tautology).
c) (~q∧q) → p: The antecedent (~q∧q) is always false (a contradiction). An implication with a false antecedent is always true. This is always true (a tautology).
d) p ∨ (p → q): If p is true, the disjunction is true. If p is false, the implication (p → q) is true, making the disjunction true. This is always true (a tautology).
e) (p∨q) ∧ (~p∧~q): This is equivalent to saying "either p or q is true, AND both p and q are false." This is always false (a contradiction).
Only option (a) can have more than one truth value.
{"correct_answer_key": "a"}
Let's analyze each option to determine which can have more than one truth value.
a: (pvq) → q If p is True and q is False, then (True v False) -> False, which is True -> False, resulting in False. If p is False and q is True, then (False v True) -> True, which is True -> True, resulting in True. Thus, this can be True or False.
b: (~p^q) → q If p is True and q is True, then (False ^ True) -> True, which is False -> True, resulting in True. If p is False and q is False, then (True ^ False) -> False, which is False -> False, resulting in True. If p is True and q is False, then (False ^ False) -> False, which is False -> False, resulting in True. If p is False and q is True, then (True ^ True) -> True, which is True -> True, resulting in True. This is always True.
c: (~q^q) → p (~q ^ q) is always False. False -> p is always True, regardless of the value of p.
d: p v (p → q) If p is True, then True v (True -> q) is always True. If p is False, then False v (False -> q) is False v True, which is True. This is always True.
e: (pvq) (~p^~q) This is equivalent to (p v q) ^ ~(p v q), which is a contradiction and always False.
Therefore, only option 'a' can have more than one truth value (True or False).
{"correct_answer_key":"a"}
Let's analyze each option to determine which can have more than one truth value. We'll use truth tables implicitly.
a) (p∨q) → q: This is a conditional statement. If q is true, the whole statement is true. If q is false, then (p∨q) must also be false for the statement to be false. If p is false and q is false, then (p∨q) is false, and the conditional is true. If p is true and q is false, then (p∨q) is true, and the conditional is false. Thus, this statement can have more than one truth value.
b) (~p∧q) → q: If q is true, the whole statement is true. If q is false, the antecedent (~p∧q) is false, and the conditional is true. Thus, this statement is always true.
c) (~q∧q) → p: The antecedent (~q∧q) is always false (a contradiction). A conditional with a false antecedent is always true. Thus, this statement is always true.
d) p ∨ (p → q): If p is true, the statement is true. If p is false, then (p → q) is true, and the statement is true. Thus, this statement is always true.
e) (p∨q) ∧ (~p∧~q): This is a conjunction. (p∨q) and (~p∧~q) are mutually exclusive. If p is true and q is false, then (p∨q) is true and (~p∧~q) is false, making the whole statement false. If p is false and q is true, then (p∨q) is true and (~p∧~q) is false, making the whole statement false. If p and q are both true, then (p∨q) is true and (~p∧~q) is false, making the whole statement false. If p and q are both false, then (p∨q) is false and (~p∧~q) is true, making the whole statement false. Thus, this statement is always false.
Therefore, only option (a) can have more than one truth value.
{"correct_answer_key": "a"}
คำถามถามว่าข้อใดต่อไปนี้สามารถมีค่าความจริงได้มากกว่า 1 ค่า ซึ่งหมายถึงข้อความที่เป็น "สัจจะนิรันดร์" (Tautology - จริงเสมอ) หรือ "ข้อขัดแย้ง" (Contradiction - เท็จเสมอ) ไม่ใช่คำตอบที่ถูกต้อง เรากำลังมองหาข้อความที่เป็น "อุบัติการณ์" (Contingency) ซึ่งค่าความจริงขึ้นอยู่กับค่าความจริงของประพจน์ย่อย (p และ q)
มาพิจารณาแต่ละตัวเลือก:
A. (p v q) → q
B. (~p ^ q) → q
C. (~q ^ q) → p
D. p v (p → q)
E. (p v q) (~p ^ ~q)
(p v q) ^ (~p ^ ~q) (และ)จากทั้งหมด ตัวเลือก A เท่านั้นที่สามารถมีค่าความจริงได้มากกว่า 1 ค่า (ทั้งจริงและเท็จ)
{"correct_answer_key":"a"}
The question asks which of the given logical expressions can have more than one truth value, meaning it's not a tautology (always true) or a contradiction (always false). Let's analyze each option:
a) (pvq) → q
b) (~p^q) → q
c) (~q^q) → p
d) p v (p → q)
e) (pvq) (~p^~q)
Based on the clear analysis of option 'a', it is the correct answer.
{"correct_answer_key":"a"}
คำถามต้องการหาประพจน์ที่สามารถมีค่าความจริงได้ทั้งจริงและเท็จ (ไม่เป็นสัจนิรันดร์หรือนิเสธสัจนิรันดร์)
a: (p v q) → q
b: (~p ^ q) → q เป็นสัจนิรันดร์ (จริงเสมอ)
c: (~q ^ q) → p เป็นสัจนิรันดร์ (จริงเสมอ) เพราะ (~q ^ q) เป็นเท็จเสมอ
d: p v (p → q) เป็นสัจนิรันดร์ (จริงเสมอ)
e: (p v q) (~p ^ ~q) รูปแบบไม่สมบูรณ์ แต่หากหมายถึง (p v q) ^ (~p ^ ~q) จะเป็นข้อขัดแย้ง (เท็จเสมอ)
ดังนั้น ข้อ a เป็นประพจน์เดียวที่สามารถมีค่าความจริงได้มากกว่า 1 ค่า
{"correct_answer_key":"a"}
This question is in Thai. Here's the translation and explanation:
Translation:
"Given that p and q are propositions, which of the following can have more than one truth value?"
Explanation:
We need to find the statement that isn't always true or always false (a tautology or contradiction). Let's analyze each option using a truth table approach (though we can often reason it out):
Therefore, only option 'a' can have more than one truth value (both true and false).
{"correct_answer_key":"a"}
The question asks which logical expression can have more than one truth value, meaning it is not a tautology (always true) or a contradiction (always false), and its truth value depends on the assignments of truth values to propositions p and q.
Option a: (p ∨ q) → q
This expression is false when p is true and q is false, and true in all other cases. Thus, it can be either true or false, depending on the truth values of p and q. It has more than one truth value.
Option b: (~p ∧ q) → q
This is always true (a tautology), regardless of p and q. It has only one truth value: true.
Option c: (~q ∧ q) → p
The sub-expression (~q ∧ q) is always false (a contradiction), making the entire implication always true. It has only one truth value: true.
Option d: p ∨ (p → q)
This simplifies to a tautology (always true) because p ∨ (¬p ∨ q) is equivalent to true ∨ q, which is always true. It has only one truth value: true.
Option e: (p ∨ q) ∧ (~p ∧ ~q)
This simplifies to (p ∨ q) ∧ ¬(p ∨ q), which is always false (a contradiction). It has only one truth value: false.
Only option a can have more than one truth value, as it is contingent (true for some assignments and false for others).
{"correct_answer_key":"a"}
The question asks which of the given logical expressions can have more than one truth value (i.e., is contingent, meaning it is neither always true nor always false). We analyze each option:
Option a: (p ∨ q) → q
This expression evaluates to false when p is true and q is false, and true otherwise. Thus, it has both true and false values, making it contingent.
Option b: (~p ∧ q) → q
This is a tautology (always true) because the antecedent (~p ∧ q) implies q regardless of p and q. It has only one truth value.
Option c: (~q ∧ q) → p
The antecedent (~q ∧ q) is always false (contradiction). An implication with a false antecedent is always true, so this is a tautology with only one truth value.
Option d: p ∨ (p → q)
Simplifying: p → q is equivalent to ~p ∨ q, so p ∨ (~p ∨ q) = (p ∨ ~p) ∨ q = true ∨ q = true. This is always true (tautology), with one truth value.
Option e: (p ∨ q) (~p ∧ ~q)
This expression is ambiguous due to the missing logical connective between the parentheses. However, interpreting it as a conjunction: (p ∨ q) ∧ (~p ∧ ~q) = (p ∨ q) ∧ ~(p ∨ q) = false (always false). Interpreting it as a disjunction: (p ∨ q) ∨ (~p ∧ ~q) = (p ∨ q) ∨ ~(p ∨ q) = true (always true). Both interpretations yield a constant truth value (either always false or always true), so it cannot have more than one truth value.
Only option a is contingent and can have more than one truth value.
{"correct_answer_key":"a"}
คำถามนี้ถามว่า "ข้อใดต่อไปนี้ สามารถมีค่าความจริงได้มากกว่า 1 ค่า" หมายถึง ประพจน์ใดที่ไม่เป็น tautology (จริงเสมอ) หรือ contradiction (เท็จเสมอ) แต่สามารถเป็นจริงหรือเท็จได้ ขึ้นกับค่าความจริงของ p และ q
มาวิเคราะห์แต่ละตัวเลือก:
a) (p v q) → q
b) (~p ^ q) → q
c) (~q ^ q) → p
d) p v (p → q)
e) (p v q) (~p ^ ~q)
ดังนั้น ตัวเลือกที่สามารถมีค่าความจริงได้มากกว่า 1 ค่า คือ ข้อ a
{"correct_answer_key":"a"}
คำอธิบาย:
ในตรรกะประพจน์ ข้อที่สามารถมีค่าความจริงได้มากกว่า 1 ค่า หมายถึงข้อที่ไม่ใช่สูตรที่เป็น "ทอทอลลี่ทรู" (true เสมอ) หรือ "ทอทอลลี่ฟอลส์" (false เสมอ) แต่เป็นสูตรที่ค่าความจริงขึ้นอยู่กับค่าของ p และ q และสามารถเป็นจริงได้ในบางกรณีและเป็นเท็จในบางกรณี
(a) ((p \lor q) \to q)
เมื่อ (p = true, q = false) จะได้ (true \to false = false)
เมื่อ (p = false, q = true) จะได้ (true \to true = true)
ดังนั้นมีค่าความจริงได้ทั้ง true และ false
(b) ((\neg p \land q) \to q)
เมื่อ (q = false) ฝั่งซ้ายเป็น false ดังนั้น (false \to false = true)
เมื่อ (q = true) ฝั่งซ้ายอาจเป็น true หรือ false แต่ผลลัพธ์จะเป็น true เสมอ
ดังนั้นสูตรนี้เป็นจริงเสมอ (ทอทอลลี่ทรู)
(c) ((\neg q \land q) \to p)
(\neg q \land q) เป็นเท็จเสมอ (contradiction)
ดังนั้นสูตรนี้เป็นจริงเสมอ (เพราะ false → p เป็นจริงเสมอ)
(d) (p \lor (p \to q))
สูตรนี้เป็นจริงเสมอ (ทอทอลลี่ทรู)
(e) ((p \lor q) (\neg p \land \neg q))
ไม่มีตัวเชื่อมชัดเจนในโจทย์ แต่ถ้าเป็น ((p \lor q) \land (\neg p \land \neg q)) จะเป็นเท็จเสมอ (contradiction)
ดังนั้นคำตอบที่มีค่าความจริงได้มากกว่า 1 ค่า คือข้อ (a)
{"correct_answer_key":"a"}
คำถามนี้ถามเกี่ยวกับประพจน์ที่สามารถมีค่าความจริงได้มากกว่า 1 ค่า ซึ่งหมายความว่าประพจน์นั้นสามารถเป็นจริงได้ในหลายกรณี (หลายชุดของค่าความจริงของ p และ q) คำตอบที่ถูกต้องคือประพจน์ที่สามารถเป็นจริงได้ในหลายกรณี
วิเคราะห์แต่ละข้อ: a) (p v q) → q: เป็นจริงในหลายกรณี เช่น ถ้า q เป็นจริง, ก็เป็นจริง; ถ้า p เป็นจริงและ q เป็นเท็จ, ก็เป็นเท็จ; แต่โดยรวมสามารถเป็นจริงได้หลายกรณี b) (~p ^ q) → q: เป็นจริงในหลายกรณี เช่น ถ้า q เป็นจริง, ก็เป็นจริง; ถ้า q เป็นเท็จ, ก็เป็นเท็จในบางกรณี c) (~q ^ q) → p: (~q ^ q) เป็นเท็จเสมอ (เพราะ q และ ~q ไม่สามารถเป็นจริงพร้อมกันได้), ดังนั้น (~q ^ q) → p เป็นจริงเสมอ (เพราะ ถ้า antecedent เป็นเท็จ, คำสั่งก็เป็นจริง) ซึ่งหมายความว่าประพจน์นี้เป็นจริงในทุกกรณี d) p v (p → q): เป็นจริงในหลายกรณี เช่น ถ้า p เป็นจริง, ก็เป็นจริง; ถ้า p เป็นเท็จ, ก็ขึ้นอยู่กับ q e) (p v q) & (~p ^ ~q): เป็นจริงในกรณีที่ p เป็นเท็จและ q เป็นเท็จเท่านั้น
จากการวิเคราะห์ ข้อ c) (~q ^ q) → p เป็นจริงในทุกกรณี เนื่องจาก antecedent (~q ^ q) เป็นเท็จเสมอ จึงทำให้ entire ประพจน์เป็นจริงในทุกกรณี ซึ่งหมายความว่าเป็นประพจน์ที่มีค่าความจริงได้มากกว่า 1 ค่า (ในทุกกรณี)
ดังนั้น คำตอบที่ถูกต้องคือ c
{"correct_answer_key":"c"}
Let's analyze each choice briefly:
a: ((p \lor q) \rightarrow q)
This implication is false only if the antecedent ((p \lor q)) is true and the consequent ((q)) is false. Thus, it can be true or false depending on the values of (p) and (q). (Multiple truth values possible)
b: ((\sim p \land q) \rightarrow q)
The antecedent ((\sim p \land q)) being true implies (q) is true, making the implication always true. If the antecedent is false, the implication is also true. Thus, this statement is always true. (Only one truth value)
c: ((\sim q \land q) \rightarrow p)
The antecedent ((\sim q \land q)) is always false, making the implication always true. (Only one truth value)
d: (p \lor (p \rightarrow q))
This statement is always true. If (p) is true, the statement is true. If (p) is false, then (p \rightarrow q) is true, making the statement true. (Only one truth value)
e: ((p \lor q)(\sim p \land \sim q))
This is a contradiction. It states that at least one of (p,q) is true and simultaneously both are false. Thus, always false. (Only one truth value)
Therefore, only option a can have more than one truth value.
{"correct_answer_key":"a"}
The question asks which proposition can have more than one truth value. In logic, a proposition that can have more than one truth value is not a tautology or a contradiction; it is a contingent statement. Let's analyze each option:
(pvq) → q: This is a conditional statement. It is not always true or always false, so it can have more than one truth value depending on the truth values of p and q.
(~p^q) → q: This is also a conditional statement. It can be true or false depending on the truth values of p and q.
(~q^q) → p: The expression (~q^q) is always false (a contradiction), so the implication (~q^q) → p is always true regardless of p, making it a tautology.
p v (p → q): This is a disjunction. It can be true or false depending on the truth values of p and q.
(pvq) (~p^~q): This expression is a conjunction of (pvq) and (~p^~q). It can be true or false depending on the truth values of p and q.
The correct answer is the one that is not a tautology or a contradiction, meaning it can have more than one truth value. In this case, options a, b, d, and e can have more than one truth value, but since the question asks for a single option, the most straightforward contingent statement is:
{"correct_answer_key":"d"}
ในการวิเคราะห์ประพจน์ที่ให้มา เราต้องพิจารณาว่าประพจน์ใดสามารถมีค่าความจริงได้มากกว่า 1 ค่า (หมายถึงสามารถเป็นจริงหรือเท็จได้ในกรณีต่าง ๆ)
จากการวิเคราะห์ข้างต้น ประพจน์ที่สามารถมีค่าความจริงได้มากกว่า 1 ค่า คือ d: p v (p → q)
{"correct_answer_key":"d"}
{"correct_answer_key":"a"}
นิพจน์ (p v q) → q จะเป็นเท็จเมื่อ p เป็นจริงและ q เป็นเท็จ แต่เป็นจริงในกรณีอื่น ๆ ดังนั้นมีค่าเป็นจริงและเท็จได้ (ข้ออื่น ๆ เป็นบทสัจนิรนัยหรือขัดแย้งเสมอ)
{"correct_answer_key":"a"}
{"correct_answer_key":"a"}
The expression ((p \lor q) \rightarrow q) simplifies to (\lnot p \lor q).
This statement is true when (p) is false or (q) is true, and false only when (p) is true and (q) is false. Hence it can take both truth values depending on the assignment of (p) and (q). All other options are either tautologies (always true) or contradictions (always false).
{"correct_answer_key":"a"}
The expression (p ∨ q) → q is not a tautology or contradiction; it evaluates to true for several truth assignments of p and q (e.g., when q is true or when both p and q are false). The other options are either always true (tautologies) or always false (contradictions).
{"correct_answer_key":"a"}
เพื่อตอบคำถามนี้ เราต้องวิเคราะห์แต่ละตัวเลือกว่าประพจน์ที่กำหนดสามารถมีค่าความจริงได้มากกว่า 1 ค่า (คือทั้ง True และ False) หรือไม่ โดยพิจารณาจากตารางค่าความจริง (truth table) ของแต่ละประพจน์
ตัวเลือก a: (p v q) → q
ประพจน์นี้ไม่ใช่ tautology หรือ contradiction เพราะค่าความจริงขึ้นอยู่กับค่า p และ q เช่น ถ้า p=True, q=False จะได้ False แต่ถ้า p=False, q=True จะได้ True ดังนั้นมีค่าความจริงได้มากกว่า 1 ค่า
ตัวเลือกอื่นๆ (b, c, d, e) จะถูกตรวจสอบในลักษณะเดียวกัน แต่จากคำถามนี้ ตัวเลือก a เป็นตัวเลือกที่ชัดเจนที่สุดที่มีค่าความจริงได้มากกว่า 1 ค่า
ดังนั้น คำตอบที่ถูกต้องคือตัวเลือก a
{"correct_answer_key":"a"}
First, the user has provided a new multiple-choice question in JSON format. It's in Thai, but I need to understand it to provide a correct response.
The question is: "กำหนดให้ p และ q เป็นประพจน์ ข้อใดต่อไปนี้ สามารถมีค่าความจริงได้มากกว่า 1 ค่า"
Translating that: "Given that p and q are propositions, which of the following can have more than one truth value?"
It's asking which logical expression can evaluate to more than one truth value, meaning it might not be a tautology or a contradiction; it could be true in some cases and false in others.
The options are:
a: (p ∨ q) → q
b: (~p ∧ q) → q
c: (~q ∧ q) → p
d: p ∨ (p → q)
e: (p ∨ q) ∧ (~p ∧ ~q)
Option e is written as "(pvq) (~p^~q)", which I assume is (p ∨ q) ∧ (~p ∧ ~q), since "^" often means AND.
I need to evaluate each one to see if it can have more than one truth value. That means, for each expression, check if it's always true, always false, or sometimes true and sometimes false.
A proposition has more than one truth value if it's contingent – true under some interpretations and false under others.
So, let's analyze each:
This is equivalent to ¬(p ∨ q) ∨ q = (¬p ∧ ¬q) ∨ q = ¬p ∨ q ∨ q (wait, better to use truth table mentally)
If q is true, then true → true is true.
If q is false, then (p ∨ false) → false = p → false, which is ¬p. So if p is true, false; if p is false, true.
So, it can be true or false depending on p and q. For example, if p=true, q=true: true
If p=true, q=false: false
So, more than one value.
The question is which one CAN have more than one value, but all might, except perhaps some are tautologies or contradictions.
I need to find which one is not a tautology or contradiction.
The question says "สามารถมีค่าความจริงได้มากกว่า 1 ค่า", which means "can have more than one truth value", implying it's not fixed to one value.
In logic, every proposition has a truth value for each assignment, but I think it means the expression as a whole can be true in some cases and false in others.
For compound propositions with variables, they all can, unless they are tautologies or contradictions.
Perhaps some are tautologies.
Let me check properly.
a: (p ∨ q) → q
As above, not a tautology; it's false when p=true and q=false.
b: (~p ∧ q) → q
(~p ∧ q) → q
This is equivalent to ¬(~p ∧ q) ∨ q = (~(~p ∧ q)) ∨ q = (p ∨ ¬q) ∨ q = p ∨ (¬q ∨ q) = p ∨ true = true
More carefully: A → B is ¬A ∨ B
So, ¬(~p ∧ q) ∨ q = [¬~p ∨ ¬q] ∨ q = [p ∨ ¬q] ∨ q = p ∨ (¬q ∨ q) = p ∨ true = true
So, b is a tautology; always true.
c: (~q ∧ q) → p
First, ~q ∧ q is always false (contradiction).
So, false → p, which is always true, because false implies anything is true.
So, c is also a tautology; always true.
d: p ∨ (p → q)
p → q is ¬p ∨ q
So, p ∨ (¬p ∨ q) = (p ∨ ¬p) ∨ q = true ∨ q = true
So, d is also a tautology; always true.
e: (p ∨ q) ∧ (~p ∧ ~q)
~p ∧ ~q is the negation of p or q.
So, (p ∨ q) ∧ (~p ∧ ~q)
~p ∧ ~q means both p and q are false.
p ∨ q is true if at least one is true.
So, if both p and q are false, p ∨ q is false, and ~p ∧ ~q is true, so false ∧ true = false
If p is true and q false, p ∨ q is true, ~p ∧ ~q is false ∧ true = false, so true ∧ false = false
If p false and q true, p ∨ q true, ~p ∧ ~q true ∧ false = false, so true ∧ false = false
If both true, p ∨ q true, ~p ∧ ~q false ∧ false = false, so true ∧ false = false
So, e is always false; a contradiction.
e is always false, so it has only one truth value: false.
The question is "can have more than one truth value", meaning it can be both true and false depending on the values.
For a: it can be true or false.
For b: always true, so only one value.
For c: always true, only one value.
For d: always true, only one value.
For e: always false, only one value.
So, only a can have more than one truth value; it can be true or false.
Is that correct? For a, as I said, when p=true q=false, it's false; otherwise true in some cases.
Yes.
The others are either always true or always false.
So, the answer should be a.
The question is: "ข้อใดต่อไปนี้ สามารถมีค่าความจริงได้มากกว่า 1 ค่า"
Which means "which of the following can have more than one truth value"
For a, yes, it can be true or false.
For b, c, d: always true, so only one value.
For e: always false, so only one value.
Therefore, only a fits.
So, the correct answer is a.
Now, I need to respond as per the initial instruction: provide a brief explanation, then respond in JSON with the "correct_answer_key".
First, brief explanation: Explain why a is the correct choice.
Then, output: {"correct_answer_key":"a"}
Make sure it's in JSON format.
กำหนดให้ p และ q เป็นประพจน์ (propositions) คำถามถามว่าข้อใดสามารถมีค่าความจริงได้มากกว่า 1 ค่า (สามารถเป็นจริงหรือเท็จได้ขึ้นอยู่กับค่าของ p และ q) โดยที่:
ดังนั้น ข้อ a เท่านั้นที่สามารถมีค่าความจริงได้มากกว่า 1 ค่า
{"correct_answer_key":"a"}
First, the user is providing another multiple-choice question in JSON format. I need to follow the same pattern as before: provide a brief explanation, then respond in JSON with the "correct_answer_key".
The question is in Thai: "กำหนดให้ p และ q เป็นประพจน์ ข้อใดต่อไปนี้ สามารถมีค่าความจริงได้มากกว่า 1 ค่า" which translates to: "Given that p and q are propositions, which of the following can have more than one truth value?"
The options are:
a: (p ∨ q) → q
b: (~p ∧ q) → q
c: (~q ∧ q) → p
d: p ∨ (p → q)
e: (p ∨ q) ∧ (~p ∧ ~q)
I need to evaluate each logical expression to see which one can take more than one truth value. That means, for which expression can the truth value vary based on the truth values of p and q.
Propositions p and q can each be true (T) or false (F), so there are 4 possible combinations: (T,T), (T,F), (F,T), (F,F).
For each option, I'll compute the truth table and see if the expression is always true, always false, or can be both true and false.
Starting with a: (p ∨ q) → q
The implication A → B is false only when A is true and B is false.
So, (p ∨ q) → q is false only if (p ∨ q) is true and q is false.
If q is false, then p ∨ q is true only if p is true. So, if p is true and q is false, then (T ∨ F) → F = T → F = F
Otherwise, it's true.
Truth table:
p=T, q=T: (T ∨ T) → T = T → T = T
p=T, q=F: (T ∨ F) → F = T → F = F
p=F, q=T: (F ∨ T) → T = T → T = T
p=F, q=F: (F ∨ F) → F = F → F = T
So, values: T, F, T, T → It can be both T and F. So, more than one value.
The question is which can have more than one value, meaning it's not a tautology or contradiction; it varies.
a can be T or F.
Now, b: (~p ∧ q) → q
Again, A → B, false only if A true and B false.
A is (~p ∧ q), B is q.
So, false only if (~p ∧ q) is true and q is false.
If q is false, then (~p ∧ q) = (~p ∧ F) = F, so A is false, and F → F = T
Truth table:
p=T, q=T: (~T ∧ T) → T = (F ∧ T) → T = F → T = T
p=T, q=F: (~T ∧ F) → F = (F ∧ F) → F = F → F = T
p=F, q=T: (~F ∧ T) → T = (T ∧ T) → T = T → T = T
p=F, q=F: (~F ∧ F) → F = (T ∧ F) → F = F → F = T
So, values: T, T, T, T → Always T. Only one value.
b is always true.
c: (~q ∧ q) → p
First, ~q ∧ q is always false, because ~q and q can't both be true.
So, A = (~q ∧ q) = F always.
Then, F → p: Implication with false antecedent is always true.
So, regardless of p, F → T = T, F → F = T
Truth table:
p=T, q=T: (~T ∧ T) → T = (F ∧ T) → T = F → T = T
p=T, q=F: (~F ∧ F) → T = (T ∧ F) → T = F → T = T
p=F, q=T: (~T ∧ T) → F = (F ∧ T) → F = F → F = T
p=F, q=F: (~F ∧ F) → F = (T ∧ F) → F = F → F = T
Always T. Only one value.
d: p ∨ (p → q)
p → q is ~p ∨ q
So, p ∨ (~p ∨ q) = (p ∨ ~p) ∨ q = T ∨ q = T
By distributive law, it's always true.
Truth table:
p=T, q=T: T ∨ (T → T) = T ∨ (T) = T ∨ T = T
p=T, q=F: T ∨ (T → F) = T ∨ (F) = T ∨ F = T
p=F, q=T: F ∨ (F → T) = F ∨ (T) = F ∨ T = T
p=F, q=F: F ∨ (F → F) = F ∨ (T) = F ∨ T = T (since F → F is T)
Always T.
e: (p ∨ q) ∧ (~p ∧ ~q)
Let's compute:
A = p ∨ q
B = ~p ∧ ~q
So, A ∧ B
~p ∧ ~q is true only when both p and q are false.
p ∨ q is true unless both are false.
If both p and q are false, p ∨ q = F, so A ∧ B = F ∧ T = F
If not both false, say p=T, q=T: p ∨ q = T, ~p ∧ ~q = F ∧ F = F, so T ∧ F = F
p=T, q=F: p ∨ q = T, ~p ∧ ~q = F ∧ T = F ∧ T = F, so T ∧ F = F
p=F, q=T: p ∨ q = T, ~p ∧ ~q = T ∧ F = F, so T ∧ F = F
p=F, q=F: p ∨ q = F, ~p ∧ ~q = T ∧ T = T, so F ∧ T = F
So, always F.
Truth table for e: Always false.
Summarizing:
a: Can be T or F (from the truth table: T, F, T, T)
b: Always T
c: Always T
d: Always T
e: Always F
The question is: which can have more than one truth value? Meaning, which is not constant; it can be true in some cases and false in others.
Only a varies: it can be true or false.
b, c, d are always true (one value: true)
e is always false (one value: false)
So, only a has more than one possible truth value.
The question says: "สามารถมีค่าความจริงได้มากกว่า 1 ค่า" which means "can have more than one truth value", implying it can take different values depending on p and q.
Yes, so a is the correct one.
In the options, e is written as "(pvq) (~p^~q)", which I assume is (p ∨ q) ∧ (~p ∧ ~q), as ^ often means ∧.
Yes.
So, the correct answer is a.
Now, respond with a brief explanation, then in JSON: {"correct_answer_key":"a"}
กำหนดให้ p และ q เป็นประพจน์ โดยคำถามถามว่าข้อใดสามารถมีค่าความจริงได้มากกว่า 1 ค่า (หมายถึงค่าความจริงสามารถเป็นจริงหรือเท็จได้ขึ้นอยู่กับค่า p และ q) เราสามารถตรวจสอบโดยใช้ตารางความจริง:
ดังนั้น ข้อ a เป็นข้อที่สามารถมีค่าความจริงได้มากกว่า 1 ค่า
{"correct_answer_key":"a"}
เพื่อตอบคำถามนี้ เราต้องวิเคราะห์แต่ละตัวเลือกว่าประพจน์ที่กำหนดสามารถมีค่าความจริงได้มากกว่า 1 ค่า (คือ เป็นได้ทั้งจริงและเท็จ ขึ้นอยู่กับค่าของ p และ q) หรือไม่ โดยประพจน์ที่เป็น tautology (จริงเสมอ) หรือ contradiction (เท็จเสมอ) จะมีค่าได้เพียงค่าเดียว
ตัวเลือก a: (p v q) → q
ประพจน์นี้ไม่ใช่ tautology เพราะมีกรณีที่เป็นเท็จได้ (เช่น p = True, q = False) ดังนั้นสามารถมีค่าความจริงได้มากกว่า 1 ค่า
ตัวเลือก b: (~p ^ q) → q
ประพจน์นี้เป็น tautology เพราะเป็นจริงเสมอ (ถ้า ~p ^ q เป็นจริง q ต้องเป็นจริงอยู่แล้ว) จึงมีค่าได้เพียงค่าเดียว
ตัวเลือก c: (~q ^ q) → p
ประพจน์นี้เป็น tautology เพราะ ~q ^ q เป็น contradiction (เท็จเสมอ) ทำให้ implication เป็นจริงเสมอ จึงมีค่าได้เพียงค่าเดียว
ตัวเลือก d: p v (p → q)
ประพจน์นี้เป็น tautology เพราะเป็นจริงเสมอ (ถ้า p เป็นจริง ผลเป็นจริง; ถ้า p เป็นเท็จ p → q เป็นจริง) จึงมีค่าได้เพียงค่าเดียว
ตัวเลือก e: (p v q) ^ (~p ^ ~q)
ประพจน์นี้เป็น contradiction เพราะ (p v q) และ (~p ^ ~q) ขัดแย้งกันเสมอ จึงมีค่าได้เพียงค่าเดียว (เท็จ)
ดังนั้น ตัวเลือกที่สามารถมีค่าความจริงได้มากกว่า 1 ค่าคือตัวเลือก a
{"correct_answer_key":"a"}
{ "correct_answer_key": "a" }
คำถามถามว่า “ข้อใดสามารถมีค่าความจริงได้มากกว่า 1 ค่า” ซึ่งหมายถึงประพจน์นั้นไม่ใช่ประพจน์แทน (tautology) และไม่ใช่ประพจน์เท็จตลอด (contradiction) จึงมีทั้ง “จริง” และ “เท็จ” ได้ตามค่าความจริงของ p และ q
ดังนั้นข้อที่สามารถมีค่าความจริงได้มากกว่า 1 ค่า คือ e
{"correct_answer_key":"e"}
ประพจน์ (pvq) → q สามารถมีค่าความจริงได้มากกว่า 1 ค่า เนื่องจากเมื่อ q เป็นจริง ประพจน์นี้จะเป็นจริงเสมอ ไม่ว่า p จะเป็นจริงหรือเท็จ และเมื่อ q เป็นเท็จ ประพจน์นี้จะเป็นเท็จเสมอ ไม่ว่า p จะเป็นจริงหรือเท็จ
{"correct_answer_key":"a"}
To solve this, let's analyze each given proposition and determine its truth value under different conditions for p and q.
First, we need to understand the given propositions and their components:
((p \lor q)) is true if (p) or (q) (or both) are true. For ((p \lor q) \to q) to be true, whenever (p \lor q) is true, (q) must be true. This is not always the case (e.g., when (p) is true and (q) is false), so let's check its truth table:
Let's briefly consider the nature of the other options:
From the steps above, we see that:
Thus, the proposition that can have more than one truth value is option a.
{"correct_answer_key":"a"}
To solve this, let's analyze each option given that p and q are propositions:
Option A is (p∨q) → q. This statement is false only when p∨q is true and q is false. This happens when p is true and q is false. So, it can have more than one truth value depending on p and q.
Option B is (~p∧q) → q. This statement is always true because if ~p∧q is true, then q must be true, making the implication true. If ~p∧q is false, the implication is still true. So, it has only one truth value.
Option C is (~q∧q) → p. Since ~q∧q is always false (because something cannot both be and not be at the same time), the implication (~q∧q) → p is always true, regardless of p. So, it has only one truth value.
Option D is p ∨ (p → q). This can be rewritten as p ∨ (~p ∨ q), which simplifies to (p ∨ ~p) ∨ q, and further to T ∨ q, which is always true. So, it has only one truth value.
Option E is (p∨q) ∧ (~p∧~q). For this to be true, both p∨q and ~p∧~q must be true. This is only possible when p and q are both false, making ~p∧~q true and p∨q false, which means this statement can only have one truth value (false), except it's constructed in a way that it's actually a contradiction for it to be true under any circumstance because if p∨q is true, ~p∧~q must be false and vice versa.
From the analysis, Option A can have more than one truth value because its truth value depends on the specific values of p and q.
{"correct_answer_key":"a"}
The question asks which of the given logical expressions can have more than one truth value.
Let's analyze each option:
Option A: ((p \lor q) \rightarrow q)
Option B: ((~p \land q) \rightarrow q)
Option C: ((~q \land q) \rightarrow p)
Option D: (p \lor (p \rightarrow q))
Option E: ((p \lor q) \land (~p \land ~q))
Given the analysis, option D is the one that can have more than one truth value.
{"correct_answer_key":"d"}
To determine which expression can have more than one true value, we need to analyze each option:
Among these, option d: p v (p → q) can have more than one true value because it is always true regardless of the values of p and q.
{"correct_answer_key":"d"}
To determine which of the given propositions can have more than one truth value, we need to analyze each option:
a) ((p \land q) \rightarrow q): This is a conditional statement. It is true unless ((p \land q)) is true and (q) is false. This can only have one truth value depending on the truth values of (p) and (q).
b) ((\neg p \land q) \rightarrow q): This is another conditional statement. It is true unless ((\neg p \land q)) is true and (q) is false. This can also only have one truth value depending on the truth values of (p) and (q).
c) ((\neg q \land q) \rightarrow p): This is a conditional statement. The antecedent ((\neg q \land q)) is always false because it is a contradiction. Therefore, the whole statement is always true regardless of the truth value of (p). This can have more than one truth value because it is always true.
d) (p \lor (p \rightarrow q)): This is a disjunction. It is true unless (p) is false and ((p \rightarrow q)) is false. This can only have one truth value depending on the truth values of (p) and (q).
e) ((p \land q) \land (\neg p \land \neg q)): This is a conjunction. It is always false because it involves a conjunction of (p \land q) and (\neg p \land \neg q), which is a contradiction. This can only have one truth value.
Therefore, the proposition that can have more than one truth value is:
{"correct_answer_key":"c"}
The expression "p v (p → q)" can have more than one truth value. The implication "p → q" is false only when p is true and q is false. In all other cases, "p → q" is true. Therefore, "p v (p → q)" is true in all cases because if p is false, the disjunction is true due to "p → q" being true, and if p is true, the disjunction is true due to p being true. Hence, option D can have multiple truth values, specifically it is always true.
{"correct_answer_key":"d"}
(p ∨ q) → q สามารถเป็นจริงหรือเท็จได้ขึ้นอยู่กับค่าความจริงของ p และ q จึงมีค่าความจริงได้มากกว่า 1 ค่า
ส่วนตัวเลือกอื่น ๆ เป็นประพจน์ที่มีค่าเป็นจริงเสมอ (tautology) หรือเท็จเสมอ (contradiction)
{ "correct_answer_key": "a" }
ความต้องการคือหาประโยคที่ไม่เป็นทั้ง “จริงเสมอ” (tautology) และไม่เป็น “เท็จเสมอ” (contradiction) นั่นคือเป็น contingent ซึ่งอาจให้ค่า True หรือ False ได้มากกว่า 1 ค่า
ตรวจสอบแต่ละตัวเลือก (p‑คือ p, q‑คือ q, ~ คือ not, v คือ or, ^ คือ and, → คือ if‑then):
a) (p v q) → q
ค่าเป็น False ได้เมื่อ p=T, q=F และเป็น True ได้ในกรณีอื่น จึงมีได้มากกว่าหนึ่งค่า
b) (~p ^ q) → q
บรรทัดฐานแสดงว่า True เสมอ (tautology)
c) (~q ^ q) → p
(~q ^ q) เท็จเสมอ → p จึง True เสมอ
d) p v (p → q)
True เสมอ
e) (p v q) ^ (~p ^ ~q)
เป็นไปไม่ได้ที่จะจริงพร้อมกัน (contradiction) จึง False เสมอ
ดังนั้นมีเพียงข้อ a เท่านั้นที่ให้ค่าความจริงได้มากกว่าหนึ่งค่า
{"correct_answer_key":"a"}
คำถามต้องการพิจารณาว่าประพจน์ผสมใดไม่ได้เป็น tautology (จริงเสมอ) และไม่ได้เป็น contradiction (เท็จเสมอ) จึงอาจให้ค่าความจริงได้ทั้งจริงและเท็จ
ตรวจทีละตัวเลือก (ใช้กฎ ~X ∨ Y แทน X→Y):
a) (p ∨ q) → q
= ¬(p ∨ q) ∨ q
= (¬p ∧ ¬q) ∨ q
= ¬p ∨ q
ถ้า q เป็นเท็จและ p เป็นจริง ผลลัพธ์เป็นเท็จ; กรณีอื่น ๆ เป็นจริง → มีได้ 2 ค่า
b) (¬p ∧ q) → q
= ¬(¬p ∧ q) ∨ q
= (p ∨ ¬q) ∨ q
= p ∨ T = T → จริงเสมอ (tautology)
c) (¬q ∧ q) → p
เนื่องจาก (¬q ∧ q) เป็นเท็จเสมอ จึงได้ F→p = T → จริงเสมอ
d) p ∨ (p → q)
= p ∨ (¬p ∨ q)
= (p ∨ ¬p) ∨ q = T ∨ q = T → จริงเสมอ
e) (p ∨ q) ∧ (¬p ∧ ¬q)
สองส่วนขัดกันโดยสิ้นเชิง → เท็จเสมอ (contradiction)
ดังนั้นข้อเดียวที่ค่าความจริงเปลี่ยนได้คือข้อ a
{"correct_answer_key":"a"}
A proposition will take on more than one truth value only if it is neither a tautology (always true) nor a contradiction (always false).
• a) (p ∨ q) → q is sometimes false (when p = T, q = F) and sometimes true, so it has both T and F in its truth table.
• b) (~p ∧ q) → q, c) (~q ∧ q) → p, and d) p ∨ (p → q) are tautologies (always true).
• e) (p ∨ q) ∧ (~p ∧ ~q) is a contradiction (always false).
Hence, only option a meets the condition.
{"correct_answer_key":"a"}
Option a, (p v q) → q, is contingent because its truth value depends on the truth values assigned to p and q. For example, when p is true and q is false, the expression evaluates to false, whereas in other cases it may be true. The other options are either tautologies (always true) or contradictions (always false), so they do not have more than one truth value.
{"correct_answer_key":"a"}
Option A, (p v q) → q, is contingent. By testing various truth assignments for p and q, we find that it can be true in some cases and false in others. In contrast, the others are either tautologies (always true) or contradictions (always false). Therefore, only option A can have more than one truth value.
{"correct_answer_key": "a"}
For option a), (p v q) → q is contingent since its truth value depends on the specific truth assignments for p and q. For example, if p is true and q is false, the antecedent (p v q) becomes true while the consequent is false, yielding a false overall value; in other cases, the formula may be true. In contrast, option b) and option d) turn out to be tautologies (always true), option c) is also always true due to its false antecedent, and option e) is a contradiction (always false). Hence only option a) can have more than one truth value.
{"correct_answer_key": "a"}
ตัวเลือก a) (p∨q)→q เมื่อ p=1, q=0 ให้ค่าเท็จ แต่กรณีอื่นให้ค่าจริง จึงมีค่าความจริงทั้งสองค่า ขณะที่ตัวเลือกอื่นเป็น tautology หรือ contradiction มีค่าเดียวตายตัว
{"correct_answer_key":"a"}
ตัวเลือก a เพียงตัวเดียวจะให้ค่าความจริงทั้ง true และ false ขึ้นกับค่า p,q ส่วนที่เหลือเป็นอสูตรหรืออัตถัตย์ (tautology/contradiction)
{"correct_answer_key":"a"}
ข้อ a) ((p\vee q)\to q) เท่ากับ (\neg p\vee q) จึงขึ้นกับค่าความจริงของ (p) และ (q) ได้ทั้งจริงหรือเท็จ ส่วนข้ออื่นๆ เป็นอัตตภาพ (tautology) หรือ ขันขัทธ์ (contradiction) ที่มีค่าเดียวตลอด
{"correct_answer_key":"a"}
ในการหาข้อใดที่สามารถมีค่าความจริงได้มากกว่า 1 ค่า เราจำเป็นต้องวิเคราะห์แต่ละข้อดังนี้:
a) ((p \lor q) \rightarrow q): ข้อนี้จะเป็นจริงเว้นแต่ (p \lor q) เป็นเท็จหรือ (q) เป็นจริง ดังนั้น มีสถานการณ์ที่สามารถทำให้ข้อนี้เป็นจริงได้มากกว่า 1 ค่า เช่น เมื่อ (q) เป็นจริง ไม่ว่า (p) จะเป็นจริงหรือเท็จ
b) ((\neg p \land q) \rightarrow q): ข้อนี้จะเป็นจริงเสมอ เพราะถ้า (\neg p \land q) เป็นจริง แสดงว่า (q) ต้องเป็นจริง ดังนั้น ข้อนี้จะเป็นจริงเสมอ ไม่ว่า (p) และ (q) จะเป็นจริงหรือเท็จอย่างไรก็ตาม
c) ((\neg q \land q) \rightarrow p): ข้อนี้จะเป็นจริงเสมอ เพราะ (\neg q \land q) เป็นเท็จเสมอ ดังนั้น ข้อสรุปใดๆ จากเงื่อนไขเท็จจะเป็นจริงเสมอ
d) (p \lor (p \rightarrow q)): ข้อนี้จะเป็นจริงเสมอ เพราะ (p \rightarrow q) เป็นจริงเสมอเมื่อ (p) เป็นเท็จ ดังนั้น ไม่ว่า (p) จะเป็นจริงหรือเท็จ ข้อนี้จะเป็นจริงเสมอ
e) ((p \lor q) \land (\neg p \land \neg q)): ข้อนี้จะเป็นเท็จเสมอ เพราะ ((p \lor q)) และ ((\neg p \land \neg q)) ไม่สามารถเป็นจริงพร้อมกันได้
จากการวิเคราะห์ข้างต้น ข้อที่สามารถมีค่าความจริงได้มากกว่า 1 ค่าคือ a, b, c, และ d แต่ถามว่า "สามารถมีค่าความจริงได้มากกว่า 1 ค่า" หมายถึงข้อใดที่ไม่จำเป็นต้องเป็นจริงเสมอ ดังนั้น ข้อที่เหมาะสมที่สุดคือ a.
{"correct_answer_key":"a"}
ในการวิเคราะห์คำถามนี้ เราต้องพิจารณาค่าความจริงของแต่ละตัวเลือก เพื่อตรวจสอบว่าข้อใดสามารถมีค่าความจริงได้มากกว่า 1 ค่า (กล่าวคือ ข้อใดสามารถเป็นจริงหรือเป็นเท็จได้ในสถานการณ์ที่แตกต่างกัน)
ตัวเลือกที่ สามารถมีค่าความจริงได้มากกว่า 1 ค่า คือ ไม่มี เพราะทุกตัวเลือกมีค่าความจริงเพียงค่าเดียวในทุกสถานการณ์
{"correct_answer_key": "none"}
คำถามนี้ต้องการหาประพจน์ที่สามารถมีค่าความจริงได้ทั้ง "จริง" และ "เท็จ" ขึ้นอยู่กับค่าของ $ p $ และ $ q $ (ไม่ใช่ทั้งสัจนิรันดร์หรือขัดแย้ง)
วิเคราะห์ตัวเลือก:
a: $(p \lor q) \rightarrow q$
b: $(\lnot p \land q) \rightarrow q$
c: $(\lnot q \land q) \rightarrow p$
d: $ p \lor (p \rightarrow q) $
e: $(p \lor q) \land (\lnot p \land \lnot q)$
สรุป: ตัวเลือกเดียวที่มีทั้งค่าจริงและเท็จคือ a
{"correct_answer_key":"a"}
{"correct_answer_key":"a"}
{"correct_answer_key":"a"}
Okay, let's tackle this logic problem. The question is in Thai, but I can read it. It says: Given that p and q are propositions, which of the following can have more than one truth value? The options are a through e.
Hmm, so the key here is to figure out which of these expressions isn't a tautology or a contradiction. Because if it's a tautology, it's always true, and if it's a contradiction, it's always false. But if it's a contingency, then it can have more than one truth value. So the question is asking which one is a contingency.
Let me go through each option one by one.
Starting with option a: (p ∨ q) → q. Let me write out the truth table for this. Let's see:
First, p and q can each be T or F. Let's consider all four combinations.
Case 1: p=T, q=T. Then (T ∨ T) is T, so T → T is T.
Case 2: p=T, q=F. (T ∨ F)=T, so T → F is F.
Case3: p=F, q=T. (F ∨ T)=T → T is T.
Case4: p=F, q=F. (F ∨ F)=F → F is T (since F→F is T? Wait, no. Wait, implication is only F when antecedent is T and consequent is F. So F→F is T. Right.
So the truth values for a are T, F, T, T. So it can be both T and F. Wait, so that would mean it's a contingency. But wait, let me check again. Wait in case 2, when p=T and q=F, the implication is F. So yes, it can be false. So option a can have multiple truth values. But let me check the other options to see if any others are also contingencies or if maybe I made a mistake.
Option b: (~p ∧ q) → q. Let's analyze this. Let's see:
First, the antecedent is (~p ∧ q). The consequent is q. So the implication is [(~p ∧ q) → q]. Let's see each case.
Case1: p=T, q=T. ~p=F. So (~p ∧ q)=F ∧ T=F. So F→T is T.
Case2: p=T, q=F. ~p=F. (~p ∧ q)=F ∧ F=F. F→F is T.
Case3: p=F, q=T. ~p=T. (T ∧ T)=T. So T→T is T.
Case4: p=F, q=F. ~p=T. (T ∧ F)=F. F→F is T.
So all cases result in T. So this is a tautology. So option b is always true, so it can't have more than one truth value. So not the answer.
Option c: (~q ∧ q) → p. The antecedent is a contradiction because ~q ∧ q is always false. So the implication is F → p. Since F→ anything is T. So regardless of p, this is always T. So this is a tautology. So option c is always true. Not the answer.
Option d: p ∨ (p → q). Let's see. Let's rewrite p → q as ~p ∨ q. So the expression becomes p ∨ (~p ∨ q). Which is equivalent to (p ∨ ~p) ∨ q, which is T ∨ q, which is T. So this is a tautology. Wait, so that's always true. So option d is a tautology. So not the answer.
Option e: (p ∨ q) (~p ∧ ~q). Wait, the original question might have a typo here. The way it's written in the JSON is "e":"(pvq) (~p^~q)". Maybe that's supposed to be a conjunction? Like (p∨q) ∧ (~p ∧ ~q)? Or maybe an implication? The user might have missed an operator. Alternatively, maybe it's written as (p∨q) (~p^~q) which could be a biconditional? But in standard logic, that's not standard. Alternatively, maybe it's a typo and should be (p∨q) ∧ (~p ∧ ~q). Let me assume that's the case. Let's check that.
So if it's (p∨q) ∧ (~p ∧ ~q). Let's see:
First, (p∨q) is true when at least one is true. (~p ∧ ~q) is true only when both are false. So the conjunction would be true only when both parts are true. But (p∨q) is true when at least one is true, but (~p ∧ ~q) requires both to be false. So the only case where both are true is when p and q are both false. Let's check:
Case1: p=T, q=T. (T∨T)=T, (~T ∧ ~T)=F. So T ∧ F=F.
Case2: p=T, q=F. (T∨F)=T, (~T ∧ T)=F. So F.
Case3: p=F, q=T. Similar to case2, F.
Case4: p=F, q=F. (F∨F)=F, (~F ∧ ~F)=T. So F ∧ T=F. Wait, so actually, in all cases, the conjunction is F. Wait, so that would make it a contradiction. Because in case4, (p∨q) is F, and (~p ∧ ~q) is T, so F ∧ T is F. So all cases give F. So that's a contradiction. So option e is always false. So it can't have more than one truth value.
Wait, but maybe I misread the operator. If the original was written as (pvq) (~p^~q) without an operator, maybe it's a typo and should be an implication? Like (p∨q) → (~p ∧ ~q). Let me check that possibility.
If it's (p∨q) → (~p ∧ ~q). Let's see:
Case1: p=T, q=T. (T∨T)=T → (F ∧ F)=F. So T→F=F.
Case2: p=T, q=F. (T∨F)=T → (F ∧ T)=F. So T→F=F.
Case3: p=F, q=T. Similar to case2, F.
Case4: p=F, q=F. (F∨F)=F → (T ∧ T)=T. F→T=T.
So in this case, the truth values would be F, F, F, T. So it can be both F and T. So that would be a contingency. But the problem is, the original question's option e is written as "(pvq) (~p^~q)", which might be missing an operator. Since the user might have intended an implication, but without knowing, it's ambiguous. However, given the options, perhaps the user made a typo and the correct operator is an implication. Alternatively, maybe it's a conjunction. But in the original problem, the user might have intended the options correctly. Let me check again.
Wait, the user's JSON for option e is written as "e":"(pvq) (~p^~q)". Maybe the space is a mistake, and it's supposed to be (p∨q) ∧ (~p ∧ ~q). But in that case, as I saw earlier, it's a contradiction. Alternatively, maybe it's (p∨q) ↔ (~p ∧ ~q). Let me check that.
If it's a biconditional: (p∨q) ↔ (~p ∧ ~q). Let's see:
Case1: p=T, q=T. LHS=T, RHS=F. So T↔F=F.
Case2: p=T, q=F. LHS=T, RHS=F. F.
Case3: p=F, q=T. Same as case2, F.
Case4: p=F, q=F. LHS=F, RHS=T. F↔T=F. Wait, so all cases are F? No, wait, in case4, F ↔ T is F. So all cases are F. So that's a contradiction again. Hmm.
Alternatively, maybe the operator is a conjunction, so the expression is (p∨q) ∧ (~p ∧ ~q). As before, that's a contradiction. So if that's the case, then option e is a contradiction. So the only options that are contingencies are a and maybe e if it's an implication. But since the user's input has e as written, perhaps the correct answer is a. Wait, but let me re-examine option a again.
Option a: (p∨q) → q. Let me recheck its truth table:
Case1: p=T, q=T → T→T=T
Case2: p=T, q=F → T→F=F
Case3: p=F, q=T → T→T=T
Case4: p=F, q=F → F→F=T
So the truth values are T, F, T, T. So it can be both T and F. So that's a contingency. So option a is a contingency. But the question is asking which can have more than one truth value, so that's a possible answer. But let me check again the other options to be sure.
Wait, option d was p ∨ (p→q). Let me recheck that. p→q is equivalent to ~p ∨ q. So p ∨ (~p ∨ q) is equivalent to (p ∨ ~p) ∨ q, which is T ∨ q, which is T. So that's a tautology. So d is always true.
So the only contingency is option a. But wait, the user's question is in Thai, and the options are labeled a to e. The question is asking which can have more than one truth value, so the answer should be a. But wait, let me check again.
Wait, but in the options given, the user's option e might have a different structure. Let me check again the original problem's option e. The user wrote "e":"(pvq) (~p^~q)". Maybe the space is a typo and it's supposed to be (p∨q) → (~p ∧ ~q). If that's the case, then as I saw earlier, that would be a contingency (since in case4 it's T, and others F). So then both a and e would be contingencies. But since the user's input has it written without an operator, maybe it's a different operator. Alternatively, maybe it's a conjunction, which is a contradiction.
Alternatively, maybe the user intended option e to be (p∨q) ∧ (~p ∧ ~q), which is a contradiction, so not a contingency. So then the only contingency is a. But let me check again the problem's wording. The question says "สามารถมีค่าความจริงได้มากกว่า 1 ค่า" which translates to "can have more than one truth value". So the answer is the one that is not a tautology or contradiction.
So based on my analysis, option a is a contingency (can be T or F), so that's the answer. But let me check again option d again to be sure.
Wait, option d: p v (p → q). Let me think of another way. Suppose p is true. Then p→q is ~p v q, so if p is true, then p→q is q. So p v (p→q) becomes T v q, which is T. If p is false, then p→q is T, so p v T is T. So regardless of p and q, it's always T. So yes, that's a tautology. So d is out.
So the only contingency is a. But wait, let me check option c again. (~q ∧ q) → p. Since ~q ∧ q is always false, so F→p is always T. So that's a tautology. So c is out.
Option b was also a tautology. So the only one left is a. So the correct answer is a. But wait, the user's options are labeled a to e, and in the first analysis, option a is a contingency. So the answer should be a. But wait, the user's first example had the correct answer as "c", but in this case, maybe the answer is a. Wait, but let me check again.
Wait, in option a: (p∨q) → q. Let me see if there's another way to simplify it. Let's see:
(p∨q) → q is equivalent to ~ (p∨q) ∨ q, which is ~p ∧ ~q ∨ q. Distribute: (~p ∨ q) ∧ (~q ∨ q). Since ~q ∨ q is T, so the whole thing is (~p ∨ q) ∧ T, which is ~p ∨ q. So (p∨q) → q is equivalent to ~p ∨ q, which is the same as p→q. Wait, that's interesting. So (p∨q) → q is equivalent to p→q. Because p→q is ~p ∨ q. So if that's the case, then the truth table for a is the same as p→q. Let's see:
p→q's truth table:
Case1: T→T=T
Case2: T→F=F
Case3: F→T=T
Case4: F→F=T
Which matches exactly what I had for option a. So yes, so (p∨q)→q is equivalent to p→q, which is a contingency. So that's correct. So the answer is a. But wait, the options given in the problem are a to e, so the correct answer key would be "a".
Wait, but the user's example had the correct answer as "c", but in this case, according to my analysis, it's a. But maybe I made a mistake here. Let me check again.
Wait, the question is asking which can have more than one truth value. So the answer is the one that is not a tautology or contradiction. So if a is a contingency, then it's the answer. But let me check again the options.
Wait, maybe I made a mistake with option e. Let me think again. If the user's option e is written as (p∨q) (~p^~q), maybe it's a conjunction, so (p∨q) ∧ (~p ∧ ~q). As I saw earlier, that's a contradiction. So it can't be. Alternatively, maybe it's an exclusive or? But that's not standard. Alternatively, maybe the user missed an operator, but given the input, I have to go with what's there.
Alternatively, maybe the user intended option e to be (p∨q) ↔ (~p ∧ ~q). Let me check that again.
(p∨q) ↔ (~p ∧ ~q).
Case1: T,T → T ↔ F → F
Case2: T,F → T ↔ F → F
Case3: F,T → T ↔ F → F
Case4: F,F → F ↔ T → F
So all cases are F. So that's a contradiction. So e is a contradiction.
Therefore, the only contingency is a. So the correct answer is a. But wait, the user's options are in Thai, but the labels are a-e. So the answer should be "a".
Wait, but let me check again the options. The user's JSON for option a is "(pvq) → q". So yes, that's the same as p→q. So that's a contingency. So the correct answer is a.
Hmm, but I need to make sure I didn't miss anything. Let me check all options again quickly:
a: contingency (T, F, T, T)
b: tautology (all T)
c: tautology (all T)
d: tautology (all T)
e: contradiction (all F)
So yes, only a is a contingency. Therefore, the correct answer is a.
Wait, but the user's example had the correct answer as "c", but in this case, the answer is a. So I need to make sure I didn't mix up the options. Let me check again the options:
The user's options are:
a: (pvq) → q
b: (~p^q) → q
c: (~q^q) → p
d: p v (p → q)
e: (pvq) (~p^~q)
Yes, so according to my analysis, a is the correct answer. So the correct answer key is "a".
The question asks which expression can have more than one truth value (i.e., is a contingency). Analyzing each option:
(p∨q) → q simplifies to p→q, which is a contingency (can be true or false).(~p^q) → q is always true (tautology).(~q^q) → p is always true (tautology).p v (p → q) simplifies to T (tautology).(p∨q) (~p^~q) (interpreted as a contradiction) is always false.Only a is a contingency, so the correct answer is a.
{"correct_answer_key":"a"}
เพื่อหาคำตอบที่ถูกต้อง เราต้องวิเคราะห์แต่ละตัวเลือก:
a. (pvq) → q: ข้อความนี้เป็นจริงเมื่อ q เป็นจริง หรือ p และ q เป็นเท็จ
b. (~p^q) → q: ข้อความนี้เป็นจริงเมื่อ q เป็นจริง หรือ p เป็นจริง
c. (~q^q) → p: ข้อความนี้เป็นจริงเสมอ เนื่องจาก ~q^q เป็นเท็จเสมอ
d. p v (p → q): ข้อความนี้เป็นจริงเมื่อ p เป็นจริง หรือ q เป็นจริง
e. (pvq) (~p^~q): ข้อความนี้เป็นเท็จเสมอ เนื่องจาก (pvq) และ (~p^~q) ไม่สามารถเป็นจริงพร้อมกันได้
จากตัวเลือกทั้งหมด ตัวเลือกที่สามารถมีค่าความจริงได้มากกว่า 1 ค่า คือ ตัวเลือก a, b, และ d
{"correct_answer_key":"d"}
Alright! Let's tackle this problem step by step. The user provided a multiple-choice question in Thai about logical statements and their possible truth values. The question is asking which of the given options can have more than one truth value.
First, I need to understand each option thoroughly. Each option is a logical expression involving p and q, which are propositions that can be either true (T) or false (F). The goal is to determine for each option whether it can have multiple truth values based on different combinations of p and q.
Starting with option a: (p ∨ q) → q. This translates to "if p or q is true, then q is true." To analyze its truth value, I'll consider all possible truth values of p and q.
From this analysis, option a can be both T and F, so it can have more than one truth value.
Moving on to option b: (~p ∧ q) → q. This translates to "if not p and q is true, then q is true." Let's break it down:
Here, option b is always T regardless of p and q's values, meaning it only has one truth value.
Next, option c: (~q ∧ q) → p. This translates to "if not q and q is true, then p is true." However, (~q ∧ q) is a contradiction because something cannot be both true and false at the same time. In classical logic, a contradiction is always false. Therefore, the implication F → p is always true, regardless of p's value. So, option c also only has one truth value.
Option d: p ∨ (p → q). Let's simplify this expression. The implication p → q is equivalent to ~p ∨ q. So, the entire expression becomes p ∨ (~p ∨ q), which simplifies to (p ∨ ~p) ∨ q. Since p ∨ ~p is a tautology (always true), the entire expression simplifies to T ∨ q, which is always true. Therefore, option d only has one truth value.
Finally, option e: (p ∨ q) ∧ (~p ∧ ~q). This translates to "p or q is true, and not p and not q is true." Breaking it down:
In all cases, option e results in F, meaning it only has one truth value.
After analyzing all options, only option a can have more than one truth value—it can be both true and false depending on the values of p and q. The other options either always result in true or always result in false, hence they don't meet the criterion of having more than one truth value.
{
"correct_answer_key": "a"
}
Let's analyze each option to determine which one can have more than one truth value.
a: (pvq) → q
b: (~p^q) → q
c: (~q^q) → p
d: p v (p → q)
e: (pvq) (~p^~q)
Therefore, option a can have more than one truth value.
{"correct_answer_key":"a"}