The number of values of r ∈ \ p, q, ~ p, ~ q\ for which ( (p ∧ q) ⇒...

Given condition is, $p ∧ q \to \left(r ∨ q\right) ∧ \left(\left(p ∧ r\right) \to q )\right) . . . . . ( 1 )$

We know conditional disjunction as,

$p \to q ≡ ~ p ∨ q$

Hence, \left(p ∧ r\right) \to q ≡ ~ \left(p ∧ r\right) ∨ q$</p><p>Using De Morgan's law</p><p>\left(p ∧ r\right) \to q ≡ ~ \left(p ∧ r\right) ∨ q ≡ \left(~ p ∨ ~ r\right) ∨ q$

Equation \left(1\right)$can be written as,</p><p>$⇒ p ∧ q \to \left(r ∨ q\right) ∧ \left(\left(~ p ∨ ~ r\right) ∨ q )\right)$</p><p>$⇒ ~ \left(p ∧ q\right) ∨ \left(r ∨ q\right) ∧ \left(\left(~ p ∨ ~ r\right) ∨ q )\right)$</p><p>$⇒ \left(~ p ∨ ~ q\right) ∨ \left(r ∨ q\right) ∧ \left(\left(~ p ∨ ~ r\right) ∨ q )\right)$</p><p>$⇒ ( ~ p ∨ \left(~ q ∨ r ∨ q\right) ∧ \left(\left(~ p ∨ ~ r\right) ∨ q )\right)$</p><p>$⇒ \left(~ p ∨ t\right) ∧ \left(\left(~ p ∨ ~ r\right) ∨ q )\right)$</p><p>$⇒ ~ p ∨ ~ r ∨ q$</p><p>For above statement to be a tautology,</p><p>$r$can be$~ p or $q$

When $r = ~ p$ or $r = q ⇒ ~ p ∨ ~ r ∨ q = t$

$∴$ Two value of $r$ are possible