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

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Question

(A) Given the expression $(p 
abla q) \Rightarrow ((p 
abla q) 
abla r)$ is a tautology.$Case-I$: If $
abla \equiv \wedge$,then $(p \wedge q) \Rig

Vedclass · Accepted Answer

$(p \Delta r) \vee q$

Vedclass · Answer

(A) Given the expression $(p 
abla q) \Rightarrow ((p 
abla q) 
abla r)$ is a tautology.$Case-I$: If $
abla \equiv \wedge$,then $(p \wedge q) \Rightarrow ((p \wedge q) \wedge r)$. This is not a tautology because if $p=T, q=T, r=F$,the expression becomes $T \Rightarrow F$,which is $F$.$Case-II$: If $
abla \equiv \vee$,then $(p \vee q) \Rightarrow ((p \vee q) \vee r)$. This is a tautology because if the antecedent $(p \vee q)$ is $T$,then the consequent $((p \vee q) \vee r)$ is also $T$.Since $
abla \equiv \vee$,we need to evaluate $(p 
abla q) \Delta r$,which is $(p \vee q) \Delta r$.If $\Delta \equiv \vee$,then $(p \vee q) \vee r \equiv (p \vee r) \vee q$,which matches option $(A)$ as $(p \Delta r) \vee q$.If $\Delta \equiv \wedge$,then $(p \vee q) \wedge r$,which does not match the options.Thus,the expression is logically equivalent to $(p \Delta r) \vee q$.

Let $\Delta, \nabla \in \{\wedge, \vee\}$ be such that $(p \nabla q) \Rightarrow ((p \nabla q) \nabla r)$ is a tautology. Then $(p \nabla q) \Delta r$ is logically equivalent to

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