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I'm trying to do a proof by contradiction, but don't quite understand how to write it down formally or how to come to an answer in this case. I'm doing a conditional statement.

The problem I'm trying to solve is "Given the premises, h ^ ~r and (h^n) --> r, show that you can conclude ~n using proof by contradiction.

I've taken the negation of both h ^ ~r and (h^n) --> r, but I'm unsure how to use these two to prove ~n

so far I've written:

(i.)~((h^n) --> r)

(ii.)~(h ^ ~r)

therefore, ~n

The hardest part I'm having is that this isn't an actual statement that I can imagine a negation of, and a step by step answer of how to do one of these proofs would be really useful, thanks!

Howard P
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    I'm pretty sure there is a sister site where this belongs, but on SO it should be a programming question and to me this is not that. – Sami Kuhmonen Jan 26 '17 at 15:44
  • I'm voting to close this question as off-topic because it's about logic / proof theory / mathematics instead of directly about programming or coding. – Pang Feb 27 '17 at 02:43

3 Answers3

1

Suppose 

~(((h ^ ~r) ^ ((h^n) --> r)) --> ~n)

Then,

~(~((h ^ ~r) ^ ((h^n) --> r)) v ~n)
=> ~(~(h ^ ~r) v ~((h^n) --> r)) v ~n)
=> ~((~h v r) v ~(~(h^n) v r)) v ~n)
=> ~((~h v r) v ((h^n) ^ ~r)) v ~n)
=> ~((~h v r) v (h ^ n ^ ~r)) v ~n)
=> ~((((~h v r v h) ^ (~h v r v n) ^ ((~h v r) v ~r)) v ~n)
=> ~(((true) ^ (~h v r v n) ^ (true)) v ~n)
=> ~((~h v r v n) v ~n)
=> ~(~h v r v n v ~n)
=> ~((~h v r) v (n v ~n))
=> ~((~h v r) v (true))
=> ~(true)
=> false //contradiction

Therefore,

((h ^ ~r) ^ ((h^n) --> r)) --> ~n
Jake
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1

Let's define:

p1 := h ^ ~r,    p2 := (h ^ n) -> r   and   q := ~n

we want to prove that p1 ^ p2 -> q.

Assume by contradiction that q=false. Then n=true. There are two cases r=true and r=false.

Case r=true

Then p1 cannot be true because ~r=false. Contradiction.

Case r=false

From p2 we deduce that (h ^ n) must be false. And given that we have assumed n=true, it must be h=false, in contradiction with p1.

Direct proof

From p1 we get h=true and r=false. Now from p2 we deduce (h ^ n) = false. And since h=true, it must be n=false, or ~n=true.

Leandro Caniglia
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1

I think the OP is probably asking about, or mis-interpretting, the structure of a proof by contradiction rather than requesting a detailed proof for the specific example.

The structure goes like this ...

  1. We've been told to assume a set of things A1, A2, ... An
  2. Let's also assume the negation of what we eventually hope to prove, i.e. ~C
  3. Do some logic that ends with any contradiction, by which we mean any statement of the form X & ~X
  4. Now we ponder what that means. Since a contradiction can never be true, there must be something wrong with at least one of our n+1 assumptions. Could be that several or all of the assumptions are false. But if any n of the assumptions are true then the remaining one cannot be true. We cannot tell at this stage which one is the problem.
  5. In this case we have been told ahead of time to accept A1, A2, ... An, and on that basis we can select the assumption of ~C as the one to be rejected.
  6. As a final step we conclude that if A1, A2, ... An are true then C must be true.
MattClarke
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