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logicproofnegation

Proof by contradiction


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!


Solution

  • 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