If the truth value of the statement pattern [ | vedclass

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(D) The given statement pattern is $[p \wedge \sim r] \rightarrow [\sim r \wedge q]$ and its truth value is $F$. An implication $A \rightarrow B$ is $

Vedclass · Accepted Answer

$\sim(r \vee q) \rightarrow \sim p$

Vedclass · Answer

(D) The given statement pattern is $[p \wedge \sim r] \rightarrow [\sim r \wedge q]$ and its truth value is $F$. An implication $A \rightarrow B$ is $F$ only when $A$ is $T$ and $B$ is $F$. So,$[p \wedge \sim r] = T$ and $[\sim r \wedge q] = F$. From $[p \wedge \sim r] = T$,we get $p = T$ and $\sim r = T$,which implies $r = F$. Since $r = F$,$\sim r = T$. Substituting $\sim r = T$ into $[\sim r \wedge q] = F$,we get $[T \wedge q] = F$,which implies $q = F$. Thus,the truth values are $p = T, q = F, r = F$. Now,check the options: $A$: $(p \vee r)$ $\rightarrow \sim r \implies (T \vee F)$ $\rightarrow T \implies T$ $\rightarrow T = T$. $B$: $(r \vee q)$ $\rightarrow \sim p \implies (F \vee F)$ $\rightarrow F \implies F$ $\rightarrow F = T$. $C$: $\sim(p \vee q)$ $\rightarrow \sim r \implies \sim(T \vee F)$ $\rightarrow T \implies \sim T$ $\rightarrow T \implies F$ $\rightarrow T = T$. $D$: $\sim(r \vee q)$ $\rightarrow \sim p \implies \sim(F \vee F)$ $\rightarrow F \implies \sim F$ $\rightarrow F \implies T$ $\rightarrow F = F$. Therefore,the statement with truth value $F$ is $\sim(r \vee q) \rightarrow \sim p$.

If the truth value of the statement pattern $[p \wedge \sim r] \rightarrow [\sim r \wedge q]$ is False,then which of the following has truth value False?

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