Question

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

• A. $(p\Delta r)\vee q$
• B. $(p\Delta r)\wedge q$
• C. $(p\wedge r)\Delta q$
• D. $(p\nabla r)\wedge q$

Answer 1

Answer: (A)

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