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Proof step 73 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, 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 74:
∀ n, bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ bits3 = [], bits2 = [], bitsand(bits2 + boolvar ', bits3 + boolvar0 ') = bitsand(bits2, bits3) + (boolvar ∧ boolvar0)'

Rule argument

Additional information

Used simplifier rules: ⊦ fillsgn(0, bits) = bits;