Proving equivalency using boolean algebra laws of logic

Question

I have a question on my exam papers relating to proving equivalences using the laws of logic, but I'm not sure how to work it out as I don't have the solution paper.

Can someone explain to me the steps involved in proving the first question (iii)? Or perhaps point me in the direction of a good tutorial on the question?

(iii) $(A\lor B)\to C \equiv (A\to C)\land (B\to C)$

(iv) $(A\to B)\to C \equiv (\neg A\to C)\land (B\to C)$

(v) $A\land (A\to B)\to B\equiv T$

Answer 1

Here is how I would go about it:

(iii) \begin{align} (A\lor B)\to C &\equiv \neg(A\lor B)\lor C\tag{$p\to q \equiv \neg p \lor q$}\\ &\equiv (\neg A\land \neg B)\lor C\tag{DeMorgan}\\ &\equiv (\neg A\lor C)\land (\neg B\lor C)\tag{distributivity}\\ &\equiv (A\to C)\land (B\to C)\tag{$p\to q \equiv \neg p \lor q$} \end{align}

(iv) \begin{align} (A\to B)\to C &\equiv \neg(A\to B)\lor C\tag{$p\to q \equiv \neg p \lor q$}\\ &\equiv \neg(\neg A\lor B)\lor C\tag{$p\to q \equiv \neg p \lor q$}\\ &\equiv (A\land \neg B)\lor C\tag{DeMorgan}\\ &\equiv (A\lor C)\land (\neg B\lor C)\tag{distributivity}\\ &\equiv (\neg A\to C)\land (B\to C)\tag{$p\to q \equiv \neg p \lor q$} \end{align}

(v) \begin{align} A \land (A\to B) &\equiv A\land (\neg A\lor B)\tag{$p\to q \equiv \neg p \lor q$}\\ &\equiv (A\land \neg A)\lor (A\land B)\tag{distributivity}\\ &\Longleftrightarrow B\equiv T\tag{$\dagger$} \end{align} $(\dagger)$ The last equivalence is the only one that is actually somewhat tricky.

Hint: Consider the possible truth values of $A$ separately and see what must be true for all of the conjunctions to be true (you will see that $B$ must be true, hence $B\equiv T$).

Answer 2

We'll look at $(iii)$.

First I will use the equivalence $(1)\;p \rightarrow q \equiv \lnot p \lor q$.

Then we apply one of DeMorgan's Laws $(2)$.

Then, we use distributivity of disjunction over conjunction $(3)$.

And for the last line $(4)$, two more applications of the equivalence to start with $(1)$.

$$\begin{align} (A\lor B) \rightarrow C & \equiv \lnot(A\lor B) \lor C \tag{1}\\ \\ & \equiv (\lnot A \land \lnot B) \lor C\tag{2}\\ \\ & \equiv (\lnot A \lor C) \land (\lnot B \lor C)\tag{3}\\ \\ &\equiv (A\rightarrow C) \land (B\rightarrow C)\tag{4}\end{align}$$