-2

I'm in the process of learning fitch proofs and I've come across one I'm having trouble setting up.

Premise: ~(A > B)
Goal: (A & ~B)

In other words, it looks something like this:

1 | ~(A > B):PR
.
.
.
2 | A:... (I've achieved this much)
.
.
.
3 | ~B:?
4 |(A & ~B):&I 2,3

So far I've got logic for concluding A, but now I struggle to conclude ~B so that I can join the two.

How should I go about doing this?

7
  • have you looked at a truth table for the material conditional? The answer should pretty much fall right out from there
    – emesupap
    Oct 3, 2022 at 0:31
  • 1
    Then you're pretty close to the finish line since then after A suppose B then you immediately have A > B which is in direct contradiction your your 1st line PR, thus by ~I you arrive at ~B... Oct 3, 2022 at 1:47
  • Thanks @DoubleKnot though I'm wondering how I can construct this A>B? I thought to construct an A>B I need a subproof above it with premise A and conclusion B?
    – Cam J
    Oct 3, 2022 at 2:24
  • 1
    Philosophically speaking as the ancient shurangama sutra hinted long ago: To speak of the false is to reveal the true, there's nothing wrong you can assume B after your A in a nested branching fashion as a first move and later arrive at a contradiction, then you can reveal the true... Oct 3, 2022 at 4:34
  • 1
    Assume B and derive A > B and this gives you the desired contradiction. Oct 3, 2022 at 6:38

1 Answer 1

1

So far I've got logic for concluding A, but now I struggle to conclude ~B so that I can join the two.

The subproofs for A and ~B are very similar in outline, with just a slight variation in details.

In each you derive a contradiction of the premise, ~(A>B), to discharge their assumptions (of ~A and B respectively) and derive their negations.

The deduction of that conditional, A>B, in each requires a Conditional Proof: Assume A so to derive B. It is just that these derivations for B are different.

In the first do so by explosion of the contradictory assumptions (ex falso quodlibet).

In the second, just reiterate : B is true under an assumption of B.

Not the answer you're looking for? Browse other questions tagged .