The number of choices of Delta in {wedge, vee | vedclass

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(B) Let $S$ be the statement $(p \Delta q) \Rightarrow ((p \Delta \sim q) \vee ((\sim p) \Delta q))$.We test each operator for $\Delta$:$1$. If $\Delt

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(B) Let $S$ be the statement $(p \Delta q) \Rightarrow ((p \Delta \sim q) \vee ((\sim p) \Delta q))$.We test each operator for $\Delta$:$1$. If $\Delta = \wedge$: $(p \wedge q) \Rightarrow ((p \wedge \sim q) \vee (\sim p \wedge q))$. This is not a tautology (e.g.,if $p=T, q=T$,then $T \Rightarrow (F \vee F) = F$).$2$. If $\Delta = \vee$: $(p \vee q) \Rightarrow ((p \vee \sim q) \vee (\sim p \vee q))$. This simplifies to $(p \vee q) \Rightarrow (T) = T$,which is a tautology.$3$. If $\Delta = \Rightarrow$: $(p$ $\Rightarrow q)$ $\Rightarrow ((p$ $\Rightarrow \sim q) \vee (\sim p$ $\Rightarrow q))$. If $p=T, q=T$,then $(T$ $\Rightarrow T)$ $\Rightarrow ((T$ $\Rightarrow F) \vee (F$ $\Rightarrow T)) = T$ $\Rightarrow (F \vee T) = T$ $\Rightarrow T = T$. If $p=T, q=F$,then $(T$ $\Rightarrow F)$ $\Rightarrow ((T$ $\Rightarrow T) \vee (F$ $\Rightarrow F)) = F$ $\Rightarrow (T \vee T) = T$. If $p=F, q=T$,then $(F$ $\Rightarrow T)$ $\Rightarrow ((F$ $\Rightarrow F) \vee (T$ $\Rightarrow T)) = T$ $\Rightarrow (T \vee T) = T$. If $p=F, q=F$,then $(F$ $\Rightarrow F)$ $\Rightarrow ((F$ $\Rightarrow T) \vee (T$ $\Rightarrow F)) = T$ $\Rightarrow (T \vee F) = T$. Thus,it is a tautology.$4$. If $\Delta = \Leftrightarrow$: $(p \Leftrightarrow q) \Rightarrow ((p \Leftrightarrow \sim q) \vee (\sim p \Leftrightarrow q))$. If $p=T, q=T$,then $(T \Leftrightarrow T)$ $\Rightarrow ((T \Leftrightarrow F) \vee (F \Leftrightarrow T)) = T$ $\Rightarrow (F \vee F) = F$. Not a tautology.Therefore,there are $2$ choices: $\vee$ and $\Rightarrow$.

The number of choices of $\Delta \in \{\wedge, \vee, \Rightarrow, \Leftrightarrow\}$,such that $(p \Delta q) \Rightarrow ((p \Delta \sim q) \vee ((\sim p) \Delta q))$ is a tautology,is

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