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.

$$
\begin{array}{ll}
&(P\vee Q) \wedge (P \Rightarrow R) \wedge (Q \Rightarrow R)\\
\equiv&\hspace{1cm}\{ \text{ by the tautology mentioned in the question }\}\\
&(P\vee Q) \wedge ( (P\vee Q) \Rightarrow R) \\
\equiv&\hspace{1cm}\{ \text{ } A \wedge (A \Rightarrow B ) \equiv A \wedge B, \text{ see below }\}\\
&(P\vee Q) \wedge R \\
\Rightarrow&\hspace{1cm}\{ \text{ } A \wedge B \Rightarrow B \text{ } \}\\
&R\\
\\
\\
&A \wedge (A \Rightarrow B)\\
\equiv&\hspace{1cm}\{ \text{ using the disjunctive definition of $\Rightarrow$ }\}\\
&A \wedge (\neg A \vee B) \\
\equiv&\hspace{1cm}\{ \text{ distribution of $\wedge$ over $\vee$ }\}\\
&(A\wedge\neg A)\vee(A \wedge B)\\
\equiv&\hspace{1cm}\{ \text{ law of excluded middle } \}\\
&\mathbf{false}\vee(A \wedge B)\\
\equiv&\hspace{1cm}\{ \text{ $\mathbf{false}$ is the unit of $\wedge$ } \}\\
&A\wedge B\\
\end{array}
$$
