Let [], [ ] [] [], [] be such that p [] q [] ((p [] q) [] r) is a tautology.

Question

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

Accepted Answer

Case-I

If $
abla$ is same as $\wedge$

Then $(p \wedge q) \Rightarrow((p \Delta q) \wedge r)$

is equivalent to $\sim(p \wedge q) \vee$ $((p \Delta q) \wedge r)$

is equivalent to $(\sim(p \wedge q) \vee(p \Delta q)) \wedge(\sim(p \wedge$ $q) \vee r)$

Which cannot be a tautology

For both $\Delta$ (i.e. $\vee$ or $\wedge$ )

Case-II

If $
abla$ is same as $v$

Then $(p \vee q) \Rightarrow((p \Delta q) \vee r)$ is equivalent to $\sim(p \vee q) \vee(p \Delta q) \vee r$

which can be a tautology if $\Delta$ is also same as $\vee$.

Hence both $\Delta$ and $
abla$ are same as $v$.

Now $(p 
abla q) \Delta r$ is equivalent to $(p \vee q \vee r)$.

Solutions · 1