Q. (Assignment 4, Problem 1(ii))

I think problem 1 (ii) is incompletely with how the parsing
is  done in the problem. Basically, I'm asking is
"is ((p -> r) -> (q -> r)) -> (~p -> q) -> r the same as
a. ((p -> r) -> (q -> r)) -> ((~p -> q) -> r)
or b. (((p -> r) -> (q -> r)) -> (~p -> q)) -> r

(I don't think we went over the unique readability of a
formula, i.e. when you get rid of parens, which way the
rule is applied first. for ex, is a -> b -> c read as
a -> (b -> c) or (a -> b) -> c?

A:

The typical convention is to view implication as right associative,

i.e.  "a -> b -> c -> d"  is the same as "a -> (b -> (c -> d))"

this makes "a -> b -> c -> d" equivalent to "(a & b & c) -> d"
(prove that as an exercise, if you like), which is more natural than
reading it as "((a -> b) -> c) -> d".

We didn't mention that in class, though, and should have added the
parentheses.

My advise for situations like that is
 a) try to find out how parentheses should be placed
 b) if you can't get an answer, state what you assume and
    then continue solving the problem based on this assumption.
    Even if we had something else in mind, we will see what you really did.