Consider the following two propositions:P_1: | vedclass

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(C) Given the proposition $p \rightarrow ((\sim p) \vee q)$ is $FALSE$.An implication $A \rightarrow B$ is $FALSE$ only when $A$ is $TRUE$ and $B$ is

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Both $P_1$ and $P_2$ are $FALSE$

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(C) Given the proposition $p \rightarrow ((\sim p) \vee q)$ is $FALSE$.An implication $A \rightarrow B$ is $FALSE$ only when $A$ is $TRUE$ and $B$ is $FALSE$.Therefore,$p$ must be $TRUE$ and $((\sim p) \vee q)$ must be $FALSE$.Since $p$ is $TRUE$,$\sim p$ is $FALSE$.For $(\sim p \vee q)$ to be $FALSE$,both $\sim p$ and $q$ must be $FALSE$. Thus,$q$ is $FALSE$.Now,evaluate $P_1$ and $P_2$ for $p = TRUE, q = FALSE$:$P_1 = \sim(p$ $\rightarrow \sim q) = \sim(T$ $\rightarrow \sim F) = \sim(T$ $\rightarrow T) = \sim(T) = FALSE$.$P_2 = (p \wedge \sim q) \wedge ((\sim p) \vee q) = (T \wedge \sim F) \wedge ((\sim T) \vee F) = (T \wedge T) \wedge (F \vee F) = T \wedge F = FALSE$.Thus,both $P_1$ and $P_2$ are $FALSE$.

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

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