Which of the following is always true? (A) ~(p → q) ≡ ~q → ~p (B) ~(p ∨ q) ≡ p ∨ ~q (C) ~(p → q) ≡ p ∧ ~q (D) ~(p ∨ q) ≡ ~p ∧ ~q
Understand the Problem
The question asks to identify which of the provided logical equivalences is always true. This involves understanding logical operators like negation (~), implication (→), disjunction (∨), conjunction (∧), and equivalence (≡), as well as De Morgan's laws.
Answer
Both (C) $\neg(p \rightarrow q) \equiv p \land \neg q$ and (D) $\neg(p \lor q) \equiv \neg p \land \neg q$ are always true.
Answer for screen readers
Both (C) and (D) are always true: (C) $\neg(p \rightarrow q) \equiv p \land \neg q$ (D) $\neg(p \lor q) \equiv \neg p \land \neg q$
Steps to Solve
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Analyze option A: $\neg(p \rightarrow q) \equiv \neg q \rightarrow \neg p$ Recall that $p \rightarrow q$ is equivalent to $\neg p \lor q$. Therefore, $\neg(p \rightarrow q)$ is equivalent to $\neg(\neg p \lor q)$, which by De Morgan's law is $p \land \neg q$. On the other hand, $\neg q \rightarrow \neg p$ is equivalent to $q \lor \neg p$. Since $p \land \neg q$ is not equivalent to $q \lor \neg p$, option A is false.
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Analyze option B: $\neg(p \lor q) \equiv p \lor \neg q$ By De Morgan's law, $\neg(p \lor q)$ is equivalent to $\neg p \land \neg q$. This is clearly not equivalent to $p \lor \neg q$, so option B is false.
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Analyze option C: $\neg(p \rightarrow q) \equiv p \land \neg q$ As noted in step 1, $\neg(p \rightarrow q)$ is equivalent to $\neg(\neg p \lor q)$, which by De Morgan's law is $p \land \neg q$. Therefore, $\neg(p \rightarrow q) \equiv p \land \neg q$ is true.
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Analyze option D: $\neg(p \lor q) \equiv \neg p \land \neg q$ By De Morgan's law, the negation of a disjunction is the conjunction of the negations. So, $\neg(p \lor q) \equiv \neg p \land \neg q$ is true.
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Verify which one is always true, using the definition of implication Option C: $\neg(p \rightarrow q) \equiv p \land \neg q$ $\neg(p \rightarrow q) \equiv \neg(\neg p \lor q) \equiv \neg \neg p \land \neg q \equiv p \land \neg q$. This holds true.
Option D: $\neg(p \lor q) \equiv \neg p \land \neg q$ This is De Morgan's law. This also holds true.
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Re-Examine options C and D using a truth table. This is because the question asks for the option that is always true. Option C: Let's analyze with a truth table. | p | q | p → q | ~(p → q) | ~q | p ∧ ~q | |---|---|-------|---------|----|-------| | T | T | T | F | F | F | | T | F | F | T | T | T | | F | T | T | F | F | F | | F | F | T | F | T | F | This equivalence holds true always. Option D: Let's analyze with a truth table. | p | q | p ∨ q | ~(p ∨ q) | ~p | ~q | ~p ∧ ~q | |---|---|-------|---------|----|----|---------| | T | T | T | F | F | F | F | | T | F | T | F | F | T | F | | F | T | T | F | T | F | F | | F | F | F | T | T | T | T | This equivalence holds true always. Both Options C and D are correct.
Both (C) and (D) are always true: (C) $\neg(p \rightarrow q) \equiv p \land \neg q$ (D) $\neg(p \lor q) \equiv \neg p \land \neg q$
More Information
De Morgan's laws are fundamental in logic and set theory. The equivalence for implication is also a crucial concept. There seem to be two valid options in this case.
Tips
A common mistake is misremembering or misapplying De Morgan's laws. Another mistake could be incorrectly negating an implication. Also, not verifying the equivalences with truth tables to ensure they always hold is another potential mistake.
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