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Proof step 77 from proof bitsand-fillsgn (previous step)

∀ n, bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ bits3 = [], bits2 = [], n = 0, fillsgn(n, bitsand(fillsgn(n - 1, bits2) + boolvar ', bits3 + boolvar0 ')) = fillsgn(n, bitsand(bits2, bits3) + (boolvar ∧ boolvar0)')

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 78:
n = m + 1, ∀ n, bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ bits3 = [], bits2 = [], n = 0, fillsgn(n, bitsand(fillsgn(m, bits2) + boolvar ', bits3 + boolvar0 ')) = fillsgn(n, bitsand(bits2, bits3) + (boolvar ∧ boolvar0)')

Rule argument

seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m] rotate: T spec: nat inst: name: elim-pred-c

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m