Is it possible to convert a conjunction of two implications into an implication of disjunctions?

Question

I want to use the following equivalence as part of a proof:

1. $$(P\implies Q ) \land (R \implies S) $$

is equivalent to

2. $$ P \lor R \implies Q \lor S$$?

Intuitively, it makes sense that $1 \implies 2$, but I'm not sure about the other direction ($2 \implies 1$). Is this correct? Why? If it is possible to prove this, which are the steps I'm missing?

Answer 1

Intuitively, if you know that $$P \lor R \implies Q \lor S$$ there is no reason to conclude that $$P \implies Q \quad\hbox{and}\quad R \implies S\ .$$ For example, it might be that $$P \implies S \quad\hbox{and}\quad R \implies Q\ .$$ So you should not expect that $(2)$ implies $(1)$. For a more formal solution, see Alexander's answer.

Answer 2

No, $(2)\implies (1)$ is false. The easiest example of that is when $S=P$ and $Q=R=\neg P$. Then $(1)\equiv\bot$ and $(2)\equiv\top$.

Answer 3

Using Implication Equivalence and other boolean algebra gives us

$\begin{split}(p\to q)\land(r\to s) &\equiv (\neg p\lor q)\land(\neg r \to s) \\&\equiv (\neg p\land\neg r)\lor(\neg p\land s)\lor(q \land \neg r)\lor(q\land s) \\&\equiv \lnot(p\lor r)\lor ((q\wedge(\lnot r\vee s))\vee(s\land(\neg p\vee q))) \\ &\equiv (p\lor r)\to ((q\land(r\to s))\lor(s\land(p\to q))) \end{split}$

Which is clearly not equivalent to $(p\lor r)\to(q\lor s)$