Associativity of logical connectives According to the precedence of logical connectives, operator $\rightarrow$ gets higher precedence than $\leftrightarrow$ operator. But what about associativity of $\rightarrow$ operator?

The implies operator ($\rightarrow$) does not have the associative property. That means that $(p \rightarrow q) \rightarrow r$ is not equivalent to $p \rightarrow (q \rightarrow r)$. Because of that, the question comes op how $p \rightarrow q \rightarrow r$ should be interpreted.

The proposition $p \rightarrow q \rightarrow r$ can be defined in multiple ways that make sense:


$(p \rightarrow q) \rightarrow r$ (left associativity)
$p \rightarrow (q \rightarrow r)$ (right associativity)
$(p \rightarrow q) \land (q \rightarrow r)$


Which one of these definitions is used?

I could not locate any book/webpage that mentions about associativity of logical operators in discrete mathematics.

Please also cite the reference (book/reliable webpage) that you use to answer my question (as I'm planning to add this to wikipedia page about 'logical connectives').

Thanks.

PS: I got this question when I saw this problem:
Check if following compound  proposition is tautology or not:

$$ \mathrm{p} \leftrightarrow (\mathrm{q} \wedge \mathrm{r}) \rightarrow \neg\mathrm{r} \rightarrow \neg\mathrm{p}$$
Some logical operators are associative: both $\wedge$ and $\vee$ are associative, as a simple check of truth tables verifies. Likewise, the biconditional $\leftrightarrow$ is associative.
However, the implication $\rightarrow$ is not associative. Compare $(p\rightarrow q)\rightarrow r$ and $p\rightarrow(q\rightarrow r)$. If all of $p$, $q$, and $r$ are false, then $p\rightarrow (q\rightarrow r)$ is true, because the antecedent is false; but $(p\rightarrow q\rightarrow r$ is false, because $r$ is false, but $p\rightarrow q$ is true. They also disagree with $p$ and $r$ are false but $q$ is true: then $p\rightarrow(q\rightarrow r)$ is true because the antecedent is false, but $(p\rightarrow q)$ is true, $r$ false, so $(p\rightarrow q)\rightarrow r$ is false.
Since they take different values at some truth assignments, the two propositions are not equivalent, so $\to$ is not associative.
Note that since it's not a binary operator, but a binary relation, associativity does not always yield the same semantics, of reducing composition pairings to a queue. Suppose $p \implies q \implies r.$ But if $r$ is false, $p \implies q$ being true does not imply $r$ is true. Likewise, $p \implies q$ can be false while $p \implies (r \implies q)$ is true, so $(p \implies (q \implies r)) \implies (p \implies q), (q \implies r)$ is not necessarily true. Finally, if $(p \implies q) \implies r$, and $p \implies (q \implies r),$ and $p \equiv F,$ we have no reason to think $q \implies r$, and $p \implies q$ can still be true, which would mean $r$ is true regardless of whether $p \implies r$.