Proving $[(P\lor Q)\land(P\to R)\land(Q\to R)]\to R$ is a tautology without using a truth table? 
$$[(P\lor Q)\land(P\to R)\land(Q\to R)]\to R\tag{1}$$

How can I prove that $(1)$ is a tautology without using a truth table?
I used the identity $$(P\to R)\land(Q\to R)\equiv(P\lor Q)\to R$$ but from there I get stuck and can't figure out where to go.

$$[(P\lor Q)\land(P\implies R)\land(Q\implies R)]\implies R$$
$$\iff\neg((P\vee Q)\wedge \neg(P\wedge\neg R)\wedge\neg(Q\wedge\neg R)\wedge\neg R)$$
$$\iff\neg((P\vee Q)\wedge\neg R\;\wedge \neg(P\wedge\neg R)\wedge\neg(Q\wedge\neg R))$$
$$\iff\neg((P\vee Q)\wedge\neg R\; \wedge(\neg P\vee R)\wedge(\neg Q\vee R))$$
$$\iff\neg((P\vee Q)\wedge\neg R \;\wedge((\neg P \wedge \neg Q)\vee R))$$
$$\iff\neg((P\vee Q)\wedge\neg R \;\wedge(\neg(P\vee Q)\vee R))$$
$$\iff\neg(\underbrace{(P\vee Q)\wedge\neg R\;}_{S} \wedge\underbrace{\neg ((P\vee Q)\wedge\neg R)}_{\neg S})$$
$$\iff\neg(S\wedge \neg S)=1$$
