The negation of the statement ( ( A ∧ ( B ∨ C ) ) ⇒ ( A ∨ B ) ) ⇒ A...

Given,

$( ( A ∧ ( B ∨ C ) ) ⇒ ( A ∨ B ) ) ⇒ A$

We know that, $X ⇒ Y ≡ ~ X ∨ Y$,

Now using the above formula we get,

$≡ \left(~ ( A ∧ ( B ∨ C ) ) ∨ ( A ∨ B )\right) ⇒ A$

$≡ ~ ( ~ ( A ∧ ( B ∨ C ) ) ∨ ( A ∨ B ) ) ∨ A$

$≡ ( A ∧ ( B ∨ C ) ) ∧ ~ ( A ∨ B ) ∨ A$

Now to solve $( A ∧ ( B ∨ C ) ) ∧ ~ ( A ∨ B )$ let us assume a universal set \left\{1, 2, 3, 4, 5, 6, 7, 8\right\}$, set$A = \left\{1, 2, 3\right\}, set $B = \left\{3, 4, 5\right\}$ and set $C = \left\{5, 6, 7\right\}</p><p>So,$A ∧ \left(B ∨ C\right)$will be \left\{3\right\}$ and ~ ( A ∨ B ) will be \left\{6, 7, 8\right\}$, </p><p>So,$( A ∧ ( B ∨ C ) ) ∧ ~ ( A ∨ B ) = ∅$</p><p>$≡ F ∨ A$</p><p>$≡ A$</p><p>$∴$Negation of statement$= ~ A