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Proof step 76 from proof bitsand-fillsgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 77:
∀ 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)')

Rule argument

Additional information

Used simplifier rules: ⊦ bits + boolvar = bits + boolvar ';% ⊦ n -1 = n - 1;