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One of the joys of hosting the PoW is the chance to show some really beautiful new ideas to you and to my students. #864 is such an example. Enjoy!

Suppose P is a property of natural numbers such that

- P[0] is true
- Whenever P[n] is true, then P[n+1] is true.
You may use the general law of implication [X and X => Y yield Y] ten times only, but -- the catch -- you are NOT allowed to use induction. What is the largest n for which you can prove P[n]?

Two examples should clarify the rules: Here is a proof of P[10].

P[0] -> P[1] -> P[2] -> P[3] -> P[4] -> P[5] -> P[6] -> P[7] -> P[8] -> P[9] -> P[10]And here is how to get to P[25]: (B) is used five times to prove an auxiliary assertion, (C), which is then used five times.

- For all n, P[n] ==> P[n+5]:
Proof:

P[n] -> P[n+1] -> P[n+2] -> P[n+3] -> P[n+4] -> P[n+5]Now use (C) five times:

P[0] -> P[5] -> P[10] -> P[15] -> P[20] -> P[25]How high a value can you get? Note that the auxiliary assertions are allowed to be complicated!

If you tell me your top value I will tell you if a higher value is possible.

Source: Michael Sheard, St. Lawrence Univ.

Thanks to Sheard and Dan Velleman and John Guilford for their comments on drafts.© Copyright 1998 Stan Wagon. Reproduced with permission.

2 October 1998