TITLE: precedence problem of multiple implication operators in logics
QUESTION [5 upvotes]: Should 
a→b→c 
be read as 
(a→b)→c 
or 
a→(b→c)?
I used a online truth table generator (http://logic.stanford.edu/intrologic/secondary/applications/babbage.html) to test and got a→(b→c) is the correct one.
But on this article it says logician use (a→b)→c  See:Boolean algebra operation precedence?
So I wondered in the field of logics, which would be the norm to read sentence with multiple implication operators such as a→b→c .

REPLY [3 votes]: In almost every context you will encounter $a \implies b \implies c$ it means
$$a \implies (b \implies c)$$
This is so common in constructive logic that it is effectively a universal standard.  The fake reason for it is that
$$a_0 \implies (a_1 \implies (a_2 \implies (\dots \implies b)))$$
is propositionally equivalent to
$$(a_0 \land a_1 \land a_2 \dots) \implies b$$
which makes it a very easy convention to work with, since most theorems have a list of conditions and 1 conclusion.  But the real reason for the convention comes from typed lambda calculus, which is the basis of constructive logic.  Suppose you have a lambda expression
$$\lambda y. \lambda x. \lambda w. V$$
and you have a predicate $T(n)$ that represents "$n$ is of the appropriate type".  Let

$A$ be $T(y)$
$B$ be $T(x)$
$C$ be $T(w)$
$V$ be $T(V)$

Then the statement that "$\lambda y. \lambda x. \lambda w. V$ is appropriately typed" is propositionally:
$$T(\lambda y. \lambda x. \lambda w. V) = (A \implies B \implies C \implies D)$$
if you associate the implication to the right as above.  The similarity between that notation and functions from $A$ to $B$ as $F: A \to B$ is apparent.  Since constructive logic is built on top of typed lambda calculus, the convention is preserved.  You will probably never encounter anywhere in modern logical publication that doesn't use this convention.