Question

Consider the following two propositions:
\(P_1 : ~ (p → ~ q)\)
\(P_2: (p ∧ ~q) ∧ ((-~p) ∨ q)\)
If the proposition \(p → ((~p) ∨ q)\) is evaluated as FALSE, then:

• A. P₁ is TRUE and P₂ is FALSE
• B. P₁ is FALSE and P₂ is TRUE
• C. Both P₁ and P₂ are FALSE
• D. Both P₁ and P₂ are TRUE

Answer 1

The Correct Option is C

To determine the truth value of the propositions \(P_1\) and \(P_2\), we first need to evaluate the given condition and logically deduce the truthfulness of \(P_1\) and \(P_2\).

The proposition \(p \to ((\sim p) \vee q)\) is given to be FALSE. In propositional logic, an implication \(A \to B\) is false only if \(A\) is true and \(B\) is false. Therefore, for the proposition to be false:

• \(p\) must be TRUE.
• \((\sim p) \vee q\) must be FALSE.

Next, since \((\sim p) \vee q\) is false, the negation of \(p\) (\(\sim p\)) is false, which implies \(p\) is true. Also, for \((\sim p) \vee q\) to be false, \(q\) must be FALSE. Hence, we have:

• \(p\) = TRUE
• \(q\) = FALSE

Now, we analyze the propositions \(P_1\) and \(P_2\):

Analyzing \(P_1: ~ (p \to \sim q)\)

The proposition \(p \to \sim q\) is false if \(p\) is TRUE and \(\sim q\) is FALSE. Since \(q\) is FALSE, \(\sim q\) becomes TRUE. Thus:

• \(p \to \sim q\) evaluates to TRUE because TRUE implies TRUE is always TRUE.

Therefore, the proposition \((p \to \sim q)\) is TRUE, so:

• \(P_1\) is FALSE as the original proposition is negated.

Analyzing \(P_2: (p \land \sim q) \land ((\sim p) \vee q)\)

For \(P_2\), breaking down step-by-step:

• \(p \land \sim q\) means TRUE \(\land\) TRUE, which is TRUE.
• \((\sim p) \vee q\) is FALSE, as \(\sim p\) is FALSE and \(q\) is FALSE. Hence, FALSE \(\vee\) FALSE is FALSE.

Combining both:

• \((p \land \sim q) \land ((\sim p) \vee q)\) results in TRUE \(\land\) FALSE, which is FALSE.

Therefore, \(P_2\) is FALSE.

Conclusion

Both \(P_1\) and \(P_2\) evaluate to FALSE. Thus, the correct answer is:

Both \(P_1\) and \(P_2\) are FALSE.