The statement (~ (p ⇔ ~ q)) ∧ q is:

$~ \left(p ⇔ ~ q\right) ∧ q$ $= \left(p ⇔ q\right) ∧ q$

Proof: Solving the given $~ \left(p ⇔ ~ q\right)$

$= ~ \left(\left(p \to ~ q\right) ∧ \left(~ q \to p\right)\right)$

$= ~ \left(\left(~ p ∨ ~ q\right) ∧ \left(q ∨ p\right)\right)$

$= ~ \left(~ \left(p ∧ q\right) ∧ \left(p ∨ q\right)\right)$

$= \left(p ∧ q\right) ∨ ~ \left(p ∨ q\right) . . . \left(1\right)$

Now, $p ⇔ q ≡ \left(p \to q\right) ∧ \left(q \to p\right)$

$= \left(~ p ∨ q\right) ∧ \left(~ q ∨ p\right)$

$= \left(~ p ∨ q\right) ∧ ~ q ∨ \left(~ p ∨ q\right) ∧ p$

$= \left(~ p ∧ ~ q\right) ∨ \left(q ∧ ~ q\right) ∨ \left(\left(~ p ∧ p\right) ∨ \left(q ∧ p\right)\right)$

$= \left(~ p ∧ ~ q\right) ∨ c ∨ \left(c ∨ \left(q ∧ p\right)\right)\left\{\text{where c is contradiction}\right\}$

$= ~ p ∧ ~ q ∨ \left(p ∧ q\right)$

$= ~ \left(p ∨ q\right) ∨ \left(p ∧ q\right) . . . \left(2\right)$

So from equation \left(1\right) \text{ and } \left(2\right)$we get</p><p>$~ \left(p ⇔ ~ q\right) ∧ q ≡ \left(p ⇔ q\right) ∧ q$</p><table border="1" cellpadding="1" cellspacing="1" style="width: 500px;"> <tbody> <tr> <td>$p$</td> <td>$q$</td> <td>$p ↔ q$</td> <td>\left(p ↔ q\right) ∧ q$ \left(p \to q\right)$</td> <td>\left(p \to q\right) ∧ q$ \left(p \to 1\right) ∧ p$</td> </tr> <tr> <td>$T$</td> <td>$T$</td> <td>$T$</td> <td>$T$</td> <td>$T$</td> <td>$T$</td> <td>$T$</td> </tr> <tr> <td>$T$</td> <td>$F$</td> <td>$F$</td> <td>$F$</td> <td>$F$</td> <td>$F$</td> <td>$F$</td> </tr> <tr> <td>$F$</td> <td>$T$</td> <td>$F$</td> <td>$F$</td> <td>$T$</td> <td>$T$</td> <td>$F$</td> </tr> <tr> <td>$F$</td> <td>$F$</td> <td>$T$</td> <td>$F$</td> <td>$T$</td> <td>$F$</td> <td>$F$</td> </tr> </tbody></table><p>$∴ \left(~ \left(p ⇔ ~ q\right)\right) ∧ q is equivalent to \left(p ⇒ q\right) ∧ p.