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