IF YOU CAN SEE THIS go to /sfa/Nuprl/Shared/Xindentation_hack_doc.html
From a sequence of moves for one fewer disks we can derive one having the largest disk on the same peg at the beginning and the end.
At:
hanoi general exists lemma1 n:, f,g:({1...n}Peg).
f(n) = g(n)
(a:, z:{a...}.
((s:({a...z}{1...n-1}Peg).
((s is a Hanoi(n-1 disk) seq on a..z ((& s(a) = f {1...n-1}Peg
((& s(z) = g {1...n-1}Peg)
( ((s:({a...z}{1...n}Peg).
((s is a Hanoi(n disk) seq on a..z & s(a) = f & s(z) = g))
1. n : 2. f : {1...n}Peg
3. g : {1...n}Peg
4. f(n) = g(n)
5. a : 6. z : {a...}
7. s:({a...z}{1...n-1}Peg).
7. s is a Hanoi(n-1 disk) seq on a..z 7. & s(a) = f {1...n-1}Peg
7. & s(z) = g {1...n-1}Peg
s:({a...z}{1...n}Peg).
s is a Hanoi(n disk) seq on a..z & s(a) = f & s(z) = g
15 steps
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IF YOU CAN SEE THIS go to /sfa/Nuprl/Shared/Xindentation_hack_doc.html