Description: Lemma for subfacp1 . In subfacp1lem6 we cut up the set of all derangements on 1 ... ( N + 1 ) first according to the value at 1 , and then by whether or not ( f( f1 ) ) = 1 . In this lemma, we show that the subset of all N + 1 derangements that satisfy this for fixed M = ( f1 ) is in bijection with N - 1 derangements, by simply dropping the x = 1 and x = M points from the function to get a derangement on K = ( 1 ... ( N - 1 ) ) \ { 1 , M } . (Contributed by Mario Carneiro, 23-Jan-2015)
Ref | Expression | ||
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Hypotheses | derang.d | |
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subfac.n | |
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subfacp1lem.a | |
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subfacp1lem1.n | |
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subfacp1lem1.m | |
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subfacp1lem1.x | |
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subfacp1lem1.k | |
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subfacp1lem3.b | |
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subfacp1lem3.c | |
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Assertion | subfacp1lem3 | |