TITLE: Is $(p \to q) \to r$ logically equivalent to $p \to (q \to r)$?
QUESTION [7 upvotes]: Is $(p \to q) \to r$ logically equivalent to $p \to (q \to r)$?
I try to simply each one, I got $\lnot ( \lnot p \lor q) \lor r$ and $\lnot p \lor ( \lnot q \lor r)$ respectively, 
then I am stuck. Does that mean they aren't equivalent? 
How do I know if expression cannot be further simply? Any tips to solve such questions?
I really appreciate your advise!

REPLY [6 votes]: Since you've gotten to:

$$\lnot ( \lnot p \lor q) \lor r\tag{1}$$ $$ \lnot p \lor ( \lnot q \lor r)\tag{2},$$

we can indeed simplify further. First, by associativity, note that $$\lnot p \lor ( \lnot q \lor r)\equiv (\lnot p \lor \lnot q) \lor r\tag{2} $$
Next, using DeMorgan's Laws,
$$\lnot (\lnot p \lor q) \lor r \equiv (p \land \lnot q) \lor r\tag{1}$$
$$(\lnot p \lor \lnot q) \lor r\equiv \lnot(p \land q) \lor r\tag{2}$$
Finally, we need only see that the clauses $$p \land \lnot q \not\equiv \lnot (p \land q)$$ to see that $(1)$ and $(2)$ are not equivalent,
Indeed, when $p$ is F(alse), $p \land \lnot q \equiv F$ and $\lnot(p \land q) \equiv T$, as we can see nicely via the corresponding truth tables (source: Wolfram Alpha):