Can you make the induction step by contradiction?

[jack476]

Assuming you've sufficiently proven your inductive basis, can you complete a proof by induction in the following manner:

Make the inductive hypothesis, assume P(n) is true for some n. Assume P(n+1) is not true. If it follows from the assumption that P(n+1) is false that P(n) must also therefore be false, contradicting the inductive hypothesis, does this mean P(n) must imply P(n+1)?

 

[andrewkirk]

Yes, that is valid.
The assumption from which you derived the contradiction was $\exists n\ P(n)\wedge\neg P(n+1)$. Deriving a contradiction from this proves its negation, which is $\forall n\neg(P(n)\wedge \neg P(n+1))$
which, by de Morgan, is equivalent to
$\forall n\ \neg P(n)\vee P(n+1)$
which is equivalent to
$\forall n\ P(n)\to P(n+1)$
which is what we need to complete the induction.

 

[jack476]

Lovely, thank you :)