If (p wedge sim q) wedge (p wedge r) rightarr | vedclass

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(C) The implication $A \rightarrow B$ is false only when $A$ is true and $B$ is false.Given $(p \wedge \sim q) \wedge (p \wedge r) \rightarrow \sim p

Vedclass · Accepted Answer

$T, F, T$

Vedclass · Answer

(C) The implication $A \rightarrow B$ is false only when $A$ is true and $B$ is false.Given $(p \wedge \sim q) \wedge (p \wedge r) \rightarrow \sim p \vee q \equiv F$.This implies $(p \wedge \sim q) \wedge (p \wedge r) \equiv T$ and $\sim p \vee q \equiv F$.From $\sim p \vee q \equiv F$,we get $\sim p \equiv F$ and $q \equiv F$,which means $p \equiv T$ and $q \equiv F$.Now substitute these into the first part: $(T \wedge \sim F) \wedge (T \wedge r) \equiv T$.$(T \wedge T) \wedge (T \wedge r) \equiv T$.$T \wedge (T \wedge r) \equiv T$.This requires $T \wedge r \equiv T$,which implies $r \equiv T$.Thus,the truth values are $p = T, q = F, r = T$.

If $(p \wedge \sim q) \wedge (p \wedge r) \rightarrow \sim p \vee q$ is false,then the truth values of $p, q$ and $r$ are respectively

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