I have been attempting to convent a prop to DNF using a group of common rules, i have applied them all but i think i should be able to get it smaller, This is what I've got so far. Thanks! $$(p \wedge (q \vee \neg p)) \rightarrow (q \wedge \neg (s \wedge r))$$ First, I remove the implication, as $p \rightarrow q$ is logically equivalent to $ \neg p \vee q$;
$$\neg (p \wedge (q \vee \neg p)) \vee (q \wedge \neg (s \wedge r))$$ Now I use the double negation rule to remove the extra nots
$$\neg (p \wedge (q \vee p)) \vee (q \wedge (s \wedge r))$$
Apply De Morgan's Laws
$$ (\neg p \wedge (q \vee p)) \wedge (\neg q \wedge (s \wedge r))$$ Use the distributive property to separate functions
$$ (\neg (p \wedge q) \vee (p \wedge p)) \wedge (\neg (q \wedge s) \vee (q \wedge r))$$
$p \wedge p$ is logically equivalent to $p$
$$ (\neg (p \wedge q) \vee (p)) \wedge (\neg (q \wedge s) \vee (q \wedge r))$$
Apply De Morgan's Laws some more
$$(\neg p \vee \neg q) \vee p) \wedge (\neg q \vee \neg s) \vee (\neg q \vee \neg r))$$
You may get DNF this way: $$ (p\wedge(1\vee\neg p))\rightarrow(q\wedge\neg(s\wedge r)) = $$ (using $x\rightarrow y = \neg x\vee y$ equality) $$ = \neg(p\wedge(q\vee\neg p))\vee(q\wedge(\neg s\vee\neg r)) = $$ (using $\neg(x\wedge y)=\neg x\vee\neg y$ equality) $$ = \neg p\vee\neg(q\vee\neg p)\vee(q\wedge(\neg s\vee\neg r)) = $$ (using $x\wedge(y\vee z) = (x\wedge y)\vee(x\wedge z)$ and $\neg(x\wedge y)=\neg x\vee\neg y$ equalities) $$ = \neg p\vee(\neg q\wedge p)\vee(q\wedge\neg s)\vee(q\wedge\neg r). $$