Question:

Consider the following two propositions: $$ P_1: \neg (p \rightarrow \neg q) $$ $$ P_2: (p \wedge \neg q) \wedge ((\neg p) \vee q) $$ If the proposition $p \rightarrow ((\neg p) \vee q)$ is evaluated as FALSE, then:

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Constructing a truth table simplifies logical evaluation.

Updated On: May 21, 2025

$ P_1 $ is TRUE and $ P_2 $ is FALSE
$ P_1 $ is FALSE and $ P_2 $ is TRUE
Both $ P_1 $ and $ P_2 $ are FALSE
Both $ P_1 $ and $ P_2 $ are TRUE

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The Correct Option is C

Approach Solution - 1

We begin by constructing a truth table for the given expressions. The statement $ p \rightarrow ((\neg p) \vee q) $ is FALSE only when $ p = T $ and $ q = F $, which gives:

\[ p \rightarrow ((\neg p) \vee q) = F \] This condition leads to $ P_1 $ and $ P_2 $ both being FALSE.

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Approach Solution -2

Step 1: Analyzing the given proposition
We are given the proposition: \[ p \rightarrow ((\neg p) \vee q) \] This is evaluated as FALSE.
Recall that a conditional proposition $ p \rightarrow q $ is false only when $ p $ is true and $ q $ is false. Therefore, for the given proposition to be false: - $ p $ must be true, - $ (\neg p) \vee q $ must be false.

Step 2: Simplifying the expression $ (\neg p) \vee q $
For $ (\neg p) \vee q $ to be false: - $ \neg p $ must be false, so $ p $ must be true (which is already given). - $ q $ must be false.
Thus, we conclude that: - $ p = \text{True} $, - $ q = \text{False} $.

Step 3: Evaluating $ P_1 $
Now, evaluate the first proposition $ P_1: \neg (p \rightarrow \neg q) $. Since $ p = \text{True} $ and $ q = \text{False} $, first evaluate $ p \rightarrow \neg q $: \[ p \rightarrow \neg q = \text{True} \rightarrow \text{True} = \text{True} \] Now, negate this result: \[ \neg (p \rightarrow \neg q) = \neg \text{True} = \text{False} \] Thus, $ P_1 $ is FALSE.

Step 4: Evaluating $ P_2 $
Next, evaluate the second proposition $ P_2: (p \wedge \neg q) \wedge ((\neg p) \vee q) $. Given that $ p = \text{True} $ and $ q = \text{False} $, evaluate each part: - $ p \wedge \neg q = \text{True} \wedge \text{True} = \text{True} $, - $ (\neg p) \vee q = \text{False} \vee \text{False} = \text{False} $.
Thus: \[ P_2 = \text{True} \wedge \text{False} = \text{False} \] Therefore, $ P_2 $ is also FALSE.

Step 5: Final Answer
Both $ P_1 $ and $ P_2 $ are FALSE:

Both $ P_1 $ and $ P_2 $ are FALSE