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.
 A: Here are the steps
$$((P \lor Q) \land (P \to R) \land (Q \to R))\to R$$
$$\equiv((\lnot(\lnot P) \lor Q) \land (P \to R) \land (Q \to R))\to R$$
$$\equiv((\lnot P \to Q) \land (Q \to R)\land (P \to R))\to R$$
$$\equiv((\lnot P \to R) \land (P \to R))\to R$$
$$\equiv((P \lor R) \land (\lnot P \lor R))\to R$$
$$\equiv(R\lor R)\to R$$
$$\equiv R\to R$$
$$\equiv \rm{true}$$
A: $$
\begin{array}{ll}
&(P\vee Q) \wedge (P \Rightarrow R) \wedge (Q \Rightarrow R)\\
\equiv&\hspace{1cm}\{ \text{ by the tautology mentioned in the question }\}\\
&(P\vee Q) \wedge ( (P\vee Q) \Rightarrow R) \\
\equiv&\hspace{1cm}\{ \text{ } A \wedge (A \Rightarrow B ) \equiv A \wedge B, \text{ see below }\}\\
&(P\vee Q) \wedge R \\
\Rightarrow&\hspace{1cm}\{ \text{ } A \wedge B \Rightarrow B \text{ } \}\\
&R\\
\\
\\
&A \wedge (A \Rightarrow B)\\
\equiv&\hspace{1cm}\{ \text{ using the disjunctive definition of $\Rightarrow$ }\}\\
&A \wedge (\neg A \vee B) \\
\equiv&\hspace{1cm}\{ \text{ distribution of $\wedge$ over $\vee$ }\}\\
&(A\wedge\neg A)\vee(A \wedge B)\\
\equiv&\hspace{1cm}\{ \text{ law of excluded middle } \}\\
&\mathbf{false}\vee(A \wedge B)\\
\equiv&\hspace{1cm}\{ \text{ $\mathbf{false}$ is the unit of $\wedge$ } \}\\
&A\wedge B\\
\end{array}
$$
A: I'll outline one fairly easy way of solving this problem without a truth table; however, the solution is not one that you obtain deduction by deduction (which is what it seems you want). 
Using DeMorgan's law and the fact that $y\to z\equiv\neg y\lor z$, we may write your implication as follows:
$$
\underbrace{(\neg p\land\neg q)}_{(1)}\lor\underbrace{(p\land\neg r)}_{(2)}\lor\underbrace{(q\land\neg r)}_{(3)}\lor r.
$$
If $r$ is true, then you're done. Thus, suppose $r$ is not true. If either $p$ or $q$ (or both) is true, then either $(2)$ or $(3)$ is true (or both). If, however, neither $p$ nor $q$ is true, then $(1)$ is true. Thus, regardless of the truth values for $p,q,r$, we must have a tautology. $\blacksquare$
A: $$[(P\lor Q)\land(P\implies R)\land(Q\implies R)]\implies R$$
$$\iff\neg((P\vee Q)\wedge \neg(P\wedge\neg R)\wedge\neg(Q\wedge\neg R)\wedge\neg R)$$
$$\iff\neg((P\vee Q)\wedge\neg R\;\wedge \neg(P\wedge\neg R)\wedge\neg(Q\wedge\neg R))$$
$$\iff\neg((P\vee Q)\wedge\neg R\; \wedge(\neg P\vee R)\wedge(\neg Q\vee R))$$
$$\iff\neg((P\vee Q)\wedge\neg R \;\wedge((\neg P \wedge \neg Q)\vee R))$$
$$\iff\neg((P\vee Q)\wedge\neg R \;\wedge(\neg(P\vee Q)\vee R))$$
$$\iff\neg(\underbrace{(P\vee Q)\wedge\neg R\;}_{S} \wedge\underbrace{\neg ((P\vee Q)\wedge\neg R)}_{\neg S})$$
$$\iff\neg(S\wedge \neg S)=1$$
A: If the formula is not a tautology, there is ( at least) one truth-value assignment in which it is false. 
The formula is a conditional which can't be false unless the consequent is false and the antecedent true. 
So, in case the whole conditional is false 


*

*R is false

*(P OR Q) , (P --> R) and ( Q --> R) are 3 true formulas ( since a conjunction is true iff alll its conjuncts are true)

*since (P--> R) is true with a false consequent ( R being false) P ( the antecedent) must be false

*since ( P OR Q) is true with a false first disjunct ( P being false), Q ( the second disjunct) must be true

*now, (Q --> R) is supposed to be true, but its antecedent ( namely : Q) is true and its consequent ( that is : R) is false ; hence a contradiction. 
So the hypothesis that the formula is false has led us to an impossible truth-value  assignment ; in other words, there is no truth value assignment that can make the formula false. 
