I am looking for the disjunctive normal form of $(A \to B) \wedge ((B \;\vee \; C) \to A)$. I know that $\neg A \vee B \equiv A \to B$. So when use this I get and the morgan law together I get:
$$(A \to B) \wedge ((B \;\vee \; C) \to A) \equiv (\neg A \vee B) \wedge (\neg (B \vee C) \vee A) \equiv (\neg A \vee B) \wedge ((\neg B \vee \neg C) \vee A) \equiv (\neg A \vee B) \wedge ( A \vee \neg B) \wedge ( A \vee \neg C)$$
Well now I don't see how to continue. Can someone maybe gives me a hint? Many thanks!
EDDIT: I try with the hint given and find: $$(A \to B) \wedge ((B \;\vee \; C) \to A) \equiv \neg(A \wedge \neg B) \wedge (\neg ((B \vee C) \wedge \neg A) \equiv (\neg A \vee B) \wedge ( \neg (B \vee C) \vee A) \equiv (\neg A \vee B \wedge \neg(B \vee C)) \vee (\neg A \vee B \wedge A) \equiv (\neg A \vee B \wedge \neg B \wedge \neg C) \vee (\neg A \vee B \wedge A)$$ But I don't think that this can be true knowing the result. Where did I again did a mistake?
Follows from your work: \begin{align} &((\lnot A\lor B)\land \lnot B\land \lnot C)\lor((\lnot A\lor B)\land A)\\ \equiv&[((\lnot A\land \lnot B)\lor (B\land \lnot B))\land \lnot C]\\ &\lor((\lnot A\land A)\lor (B\land A))\tag*{Distributive laws}\\ \equiv&[((\lnot A\land \lnot B)\lor \bot)\land \lnot C]\lor(\bot\lor (B\land A))\tag*{Negation law}\\ \equiv&(\lnot A\land \lnot B\land\lnot C)\lor(B\land A)\tag*{Identity law} \end{align}