Let Delta, nabla in {wedge, vee} be such that | vedclass

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(C) To determine which combination results in a tautology,we evaluate the truth table for each case:$1$. For $\Delta=\wedge, 
abla=\vee$: The express

Vedclass · Accepted Answer

$\Delta=\vee, 
abla=\vee$

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(C) To determine which combination results in a tautology,we evaluate the truth table for each case:$1$. For $\Delta=\wedge, 
abla=\vee$: The expression is $(p ightarrow q) \wedge (p \vee q)$. If $p=T, q=F$,then $(T ightarrow F) \wedge (T \vee F) = F \wedge T = F$. Not a tautology.$2$. For $\Delta=\vee, 
abla=\wedge$: The expression is $(p ightarrow q) \vee (p \wedge q)$. If $p=T, q=F$,then $(T ightarrow F) \vee (T \wedge F) = F \vee F = F$. Not a tautology.$3$. For $\Delta=\vee, 
abla=\vee$: The expression is $(p ightarrow q) \vee (p \vee q)$. - If $p=T, q=T$: $T \vee T = T$- If $p=T, q=F$: $F \vee T = T$- If $p=F, q=T$: $T \vee T = T$- If $p=F, q=F$: $T \vee F = T$Since all values are $T$,this is a tautology.$4$. For $\Delta=\wedge, 
abla=\wedge$: The expression is $(p ightarrow q) \wedge (p \wedge q)$. If $p=F, q=F$,then $T \wedge F = F$. Not a tautology.Thus,the correct option is $\Delta=\vee, 
abla=\vee$.

Let $\Delta, \nabla \in \{\wedge, \vee\}$ be such that $( p \rightarrow q ) \Delta ( p \nabla q )$ is a tautology. Then

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