tl;dr
our ways of proving things is reduced, but our interpretation of the meaning is richer.
Consider this: The law of excluded middle says that for all statement A, we have
A∨(¬A)
>Either 'A is true' or '(not A) is true'
or put differently,
>Either 'A is true' or 'A is false'
This doesn't hold in constructive mathematics.
Now if ⊥ denotes absurdum, one generally captures the sentence "A leads to a contradition" as
A→⊥
and one may always write A→⊥ instead of
¬A
The law of excluded middle is then written as
A∨(A→⊥)
>Either A holds, or A leads to a contradiction.
This doesn't hold constructively. It's not true that
>Either we have an argument for A, or we have an argument for A leading to a contradition.
This is because not having an arugment for A isn't counted as having an argument for a contradition
(While classically, this is just postulated by this very sentence A∨(¬A))
But consider this. There is of course the following statement that always holds
(P∧(P→B))→B
>If P and we have that (P implies B), then B
And in particular for P=A→B, the reads
((A→B)∧((A→B)→B))→B
>If (A implies B) and we have that (A implies B) implies B, then B
(if you know type thery, the constructive proof of the above is λx.(π2x)(π1x), where π are the left and right projections)
And we have
((A→⊥)∧((A→⊥)→⊥))→⊥
>If we have an argument that (A leads to a contradition) AND if having an argument that (A leads to a contradition) gives us a contradition, then we indeed have a contradition.
The above sentence doesn't give a way to conclude A→⊥ classically (i.e. ¬A) but instead ...→⊥ where ... is some convoluted expression.
However, the above is
¬(¬A∧¬(¬A))
And now CLASSICALLY we have ¬(P∧Q)↔(¬P∨¬Q) and P↔¬¬P and so this has the classical meaning
A∨(¬A)
!
We can't prove A∨(¬A) constructively, but this is classically equivalent to a statement we can prove constructively.