Prove $(R \implies W) \wedge (R \implies \neg S) = R \implies (\neg S \wedge W)$

Question

Prove $(R \implies W) \wedge (R \implies \neg S) = R \implies (\neg S \wedge W)$.

Here is my work so far: $$(R \implies W) \wedge (R \implies \neg S) \\ \equiv ((\neg R \vee W) \wedge (\neg R \vee \neg S) \\ \equiv ((\neg R\ \vee W)\wedge \neg R) \vee ((\neg R\ \vee W)\wedge \neg S) \\ \equiv (\neg R) \vee ((\neg R\ \vee W)\wedge \neg S) \\ \equiv (\neg R) \vee ((\neg R \wedge \neg S) \vee (\neg S \wedge W))$$ This is the part I do not understand, because if I distribute I seem to arrive at a never ending loop.

Answer 1

The distribution you need is simply to distribute the common term out of the two conjuncts. You should recognise that $x\lor(y\land z)$ distributes to $(x\lor y)\land(x\lor z)$, so simply flip things around.

$$(\lnot R\lor W)\land(\lnot R\lor\lnot S)~~\equiv~~ \lnot R\lor(W\land\lnot S)$$