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

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Question

$Case-I$ If $\Delta \equiv 
abla \equiv \wedge$

$(p \wedge q) ightarrow((p \wedge q) \wedge r)$

it can be false if $r$ is false,

so not a

Vedclass · Accepted Answer

$( p \Delta r ) \vee q$

Vedclass · Answer

$Case-I$ If $\Delta \equiv 
abla \equiv \wedge$

$(p \wedge q) ightarrow((p \wedge q) \wedge r)$

it can be false if $r$ is false,

so not a tautology

$Case-II$ If $\Delta \equiv 
abla \equiv \vee$

$( p \vee q ) ightarrow(( p \vee q ) \vee r ) \equiv$ tautology

then $(p \vee q) \vee r \equiv(p \Delta r) \vee q$

$Case-III$ if $\Delta=\vee, 
abla=\wedge$

then $( p \wedge q ) ightarrow\{( p \vee q ) \wedge r \}$

Not a tautology

$($ Check $p ightarrow T , q ightarrow T , r ightarrow F )$

$Case-IV$ if $\Delta=\wedge, 
abla=\vee$

$(p \wedge q) ightarrow\{(p \wedge q) \vee r\}$

Not a tautology

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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