The number of choices for Δ ∈ \ ∧ , ∨ , ⇒ , ⇔\ , such that (p ) ⇒...

If \left(\left(p\Delta ~ q\right) ∨\left(\left(~ p\right)\Deltaq\right)\right)$will be always true then the given statement \left(p\Deltaq\right) ⇒$\left(\left(p\Delta ~ q\right) ∨ \left(\left(~ p\right)\Deltaq\right)\right)will be a tautology.</p><p>Now for \left(\left(p\Delta ~ q\right) ∨ \left(\left(~ p\right)\Deltaq\right)\right)$to be true,$\Delta$should be$∨ \text{or} ⇒$</p><p>As \left(\left(p ∨ ~ q\right) ∨ \left(\left(~ p\right) ∨ q\right)\right) ≡ t; \left(\left(p ⇒ ~ q\right) ∨ \left(\left(~ p\right) ⇒ q\right)\right) ≡ t$