Description: Lemma for wilth : induction step. The "hand proof" version of this theorem works by writing out the list of all numbers from 1 to P - 1 in pairs such that a number is paired with its inverse. Every number has a unique inverse different from itself except 1 and P - 1 , and so each pair multiplies to 1 , and 1 and P - 1 == -u 1 multiply to -u 1 , so the full product is equal to -u 1 . Here we make this precise by doing the product pair by pair.
The induction hypothesis says that every subset S of 1 ... ( P - 1 ) that is closed under inverse (i.e. all pairs are matched up) and contains P - 1 multiplies to -u 1 mod P . Given such a set, we take out one element z =/= P - 1 . If there are no such elements, then S = { P - 1 } which forms the base case. Otherwise, S \ { z , z ^ -u 1 } is also closed under inverse and contains P - 1 , so the induction hypothesis says that this equals -u 1 ; and the remaining two elements are either equal to each other, in which case wilthlem1 gives that z = 1 or P - 1 , and we've already excluded the second case, so the product gives 1 ; or z =/= z ^ -u 1 and their product is 1 . In either case the accumulated product is unaffected. (Contributed by Mario Carneiro, 24-Jan-2015) (Proof shortened by AV, 27-Jul-2019)
Ref | Expression | ||
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Hypotheses | wilthlem.t | |
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wilthlem.a | |
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wilthlem2.p | |
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wilthlem2.s | |
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wilthlem2.r | |
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Assertion | wilthlem2 | |