If {(p wedge sim q) wedge (p wedge r)} righta | vedclass

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(D) conditional statement $A \rightarrow B$ is $False$ if and only if $A$ is $True$ and $B$ is $False$. Here,$A = \{(p \wedge \sim q) \wedge (p \wedge

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$T, F, T$

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(D) conditional statement $A \rightarrow B$ is $False$ if and only if $A$ is $True$ and $B$ is $False$. Here,$A = \{(p \wedge \sim q) \wedge (p \wedge r)\}$ and $B = (\sim p \vee q)$. For $B = (\sim p \vee q)$ to be $False$,both $\sim p$ and $q$ must be $False$. This implies $p = True$ and $q = False$. Now,substitute these values into $A$: $A = \{(T \wedge \sim F) \wedge (T \wedge r)\} = \{(T \wedge T) \wedge (T \wedge r)\} = \{T \wedge (T \wedge r)\} = (T \wedge r)$. For $A$ to be $True$,$r$ must be $True$. Therefore,the truth values are $p = True, q = False, r = True$.

If ${(p \wedge \sim q) \wedge (p \wedge r)} \rightarrow (\sim p \vee q)$ has a truth value of $False$,then the truth values of the statements $p, q, r$ are respectively:

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