Why are "$(P ⇒ Q) ⇒ R$" and "$P ⇒ (Q ⇒ R)$" not logically equivalent?

Question

According to the order of precedence, why are the following not logically equivalent? $$(P ⇒ Q) ⇒ R$$ $$P ⇒ (Q ⇒ R)$$ I am confused about where brackets fit into the order of precedence.

Answer 1

Suppose $P=Q=R=0$. Then $$P\Rightarrow(Q\Rightarrow R)=0\Rightarrow1=1$$ but $$(P\Rightarrow Q)\Rightarrow R=1\Rightarrow 0=0$$

Answer 2

Recall that $f\Rightarrow X$ is always true and $t\Rightarrow X$ is equivalent to $X$.

Now consider the case that $P$ is false. Then $P\Rightarrow(Q\Rightarrow R)$ is $f\Rightarrow(\ldots)$, hence true. But $(P\Rightarrow Q)\Rightarrow R$ is $(f\Rightarrow Q)\Rightarrow R$, is $t\Rightarrow R$, is $R$. As $R$ is not in generally true when $P$ is false, the two statements cannot be equivalent.

Answer 3

One has $(P\Rightarrow Q) \Rightarrow R = (\neg P \lor Q) \Rightarrow R = \neg (\neg P \lor Q) \lor R = P \land \neg Q \lor R$,

and $P\Rightarrow (Q \Rightarrow R) = P \Rightarrow (\neg Q \lor R) = \neg P \lor (\neg Q \lor R) = \neg P \lor \neg Q \lor R$.