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Suppose, that $P(k+1) \Rightarrow P(k)$ for all positive integers $k$. How would I prove $P(n)$ is true?

I am getting confused since this is going the 'other way'. Usually $P(k)\Rightarrow P(k+1)$

Brian Fitzpatrick
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atherton
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    For a non-speculative answer, you should state the full problem. There are instances of "downward" induction. – André Nicolas Mar 03 '14 at 06:17
  • Hi André, this is the full problem. Would there be a generalized way to prove "downward" induction? – atherton Mar 03 '14 at 06:23
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    I had not noticed that you had put part of the question in the long title. – André Nicolas Mar 03 '14 at 06:28
  • Related: http://math.stackexchange.com/questions/97350/how-does-backwards-induction-work-to-prove-a-property-for-all-naturals and http://math.stackexchange.com/questions/190518/induction-proof-s-contains-powers-of-2-and-predecessors-implies-s-bf-n (You can find more related questions among linked questions there.) – Martin Sleziak Mar 04 '14 at 08:36

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If $P(n)$ is sometimes false, let $a$ be a positive integer such that $P(a)$ is false. Let $2^n$ be the smallest power of $2$ (or any power of $2$) which is $\ge a$. Let $k$ be the largest positive integer $\le 2^n$ such that $P(k)$ is false. Note that $k\lt 2^n$.

It follows that $P(k+1)$ is true, which implies that $P(k)$ is true, and we have reached a contradiction.

André Nicolas
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