Prove $(p \lor r) \to (\lnot r \to p)$ Prove 
$(p \lor r) \to (\lnot r \to p)$ is a tautology.
I have tried to prove that by using truth table, but is there any another way to prove it?
Thanks in advanced!
 A: Another way to prove it it a tautology is to simply 'prove' the statement is true under no new given assumptions.
$$\begin{align} p\lor r &\implies r\lor p\\
&\implies \neg r \to p\end{align}$$
You can also assume it is false and reach a contradiction.
Suppose the statement is false. Then $p\lor r$ is true, but $\neg r \to p$ is false. Since $p\lor r$ is true, then $p$ is true or $r$ is true.
If $p$ is true, then clearly $\neg r\to p$ is true, contradiction.
If $r$ is true, then $\neg r$ is false and so $\neg r\to p$ is true, contradiction.
Therefore $\neg r\to p$ is true and hence $p\lor r\to (\neg r\to p)$.
A: 
The proof can be made entirely mechanical using the identity $(a\to b)=(\lnot a\lor b)$ for every $a$ and $b$ and the fact that $a\to a$ is a tautology for every $a$.

Let $q=(p \lor r)$, $s=(\lnot r \to p)$ and $t=q\to s$.
Then $s=(r\lor p)$ hence $t=(p \lor r)\to(r\lor p)$, which is a tautology since $(p \lor r)=(r\lor p)$.
A: Assume 'p' false.  So, (p $\lor$ r)=r (where the valuation of sentences comes as assumed).  
Thus, [(p∨r)→(¬r→p)]=[r→(¬r→p)].  '[r→(¬r→p)]' qualifies as a tautology, so [(p∨r)→(¬r→p)] holds as true when 'p' is false.
Assume 'p' true.  Thus, '(¬r→p)' qualifies as true, and so does '[(p∨r)→(¬r→p)]'.
By the principle of bivalence, it follows that '[(p∨r)→(¬r→p)]' qualifies as a tautology.
