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(A) Given: $p = T, q = T, r = F, s = F$. For $a: \sim(p \wedge \sim r) \vee(\sim q \vee s)$ Substitute the values: $\sim(T \wedge \sim F) \vee(\sim T

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$F, F$

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(A) Given: $p = T, q = T, r = F, s = F$. For $a: \sim(p \wedge \sim r) \vee(\sim q \vee s)$ Substitute the values: $\sim(T \wedge \sim F) \vee(\sim T \vee F)$ $= \sim(T \wedge T) \vee(F \vee F)$ $= \sim(T) \vee(F)$ $= F \vee F = F$. For $b: (p \vee s) \leftrightarrow(q \wedge r)$ Substitute the values: $(T \vee F) \leftrightarrow(T \wedge F)$ $= (T) \leftrightarrow(F)$ Since the truth values are different,the biconditional statement is $F$. Therefore,the truth values of $a$ and $b$ are $F, F$.

Let $a: \sim(p \wedge \sim r) \vee(\sim q \vee s)$ and $b: (p \vee s) \leftrightarrow(q \wedge r)$. If the truth values of $p$ and $q$ are $T$ and that of $r$ and $s$ are $F$,then the truth values of $a$ and $b$ are respectively...

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