Negation of p ∧ (q ∧ ~ (p ∧ q)) is

Given,

Expression $p ∧ \left(q ∧ ~ \left(p ∧ q\right)\right)$

So, negation of the above expression will be,

$~ \left[p ∧ \left(q ∧ ~ \left(p ∧ q\right)\right)\right]$

Now using demorgan's law $~ \left(A ∧ B\right) ≡ ~ A ∨ ~ B \text{ and } ~ \left(A ∨ B\right) ≡ ~ A ∧ ~ B$ we get,

$~ \left[p ∧ \left(q ∧ ~ \left(p ∧ q\right)\right)\right]$

$≡ ~ \left[p ∧ \left(q ∧ \left(~ p ∨ ~ q\right)\right)\right]$

$≡ ~ \left[p ∧ \left(\left(q ∧ ~ p\right) ∨ \left(q ∧ ~ q\right)\right)\right]$

$≡ ~ \left[p ∧ \left(\left(q ∧ ~ p\right) ∨ F\right)\right]$

$≡ ~ \left[p ∧ \left(q ∧ ~ p\right)\right]$

$≡ ~ p ∨ ~ \left(q ∧ ~ p\right)$

$≡ ~ p ∨ \left(~ q ∨ p\right)$

$≡ \left(~ p ∨ ~ q\right) ∨ p$

$≡ ~ \left(p ∧ q\right) ∨ p$