Is $(p \to q) \to r$ logically equivalent to $p \to (q \to r)$? 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!
 A: No, compare $(p\implies p)\implies r$, which is equivalent to $r$, and $p\implies (p\implies r)$, which is equivalent to $p\implies r$.
A: It is not. Suppose $p$, $q$, $r$ are false. Then $p \rightarrow q$ and $q \rightarrow r$ are true, so $(p\rightarrow q)\rightarrow r$ is false and $p\rightarrow (q\rightarrow r)$ is true.
A: 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):


A: Good question!
No.
The second statement, namely $$p \rightarrow (q \rightarrow r)$$ can be read: "In the presence of $p$, we have that $q$ implies $r$."
Turns out this is the same as saying: "In the presence of both $p$ and $q$, we have $r$." That is
$$(p \wedge q) \rightarrow r.$$
The first statement is a little harder put words to. In the comments, Lord Farin suggests the phrasing: "If $p$ necessitates $q$, then $r$."
A bit of symbol pushing shows its equivalent to:
$$(p \vee r) \wedge (q \rightarrow r)$$
which might shed some light on its meaning. Substituting in $q = \bot$ (a contradiction) we obtain 
$$(p \vee r).$$
So if $p$ necessitates a contradiction implies $r$, then either $p$ or $r$. And vice versa.
A: The second formula is "if $p$, then: if $q$ then $r$" which is equivalent to "if $p$ and $q$, then $r$". So it is equivalent to $$(p\land q) \implies r.$$
On the other hand, the first is "if $p$ implies $q,$ then $r$". As we know any  implication "$p$ implies $q$" is equivalent to "not $p$ or $q$", so our formula is "if not $p$ or $q$, then $r$", which written with symbols is: $$(\neg p \lor q) \implies r.$$
Because of course $\neg p \lor q \not\equiv p \land q$, the two formulas are not equivalent.
