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Proof step 14 from proof bitsand-fillsgn (previous step)

bits ≠ [] ∧ bits0 = true ', # bits0 = 1, ∀ n, bits, bits1. # bits1 < 1 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(fillsgn(n, bits), bits0)) = fillsgn(n, bits), bits0 = [], bits = [], # bits = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
∀ n, bits, bits1. # bits1 < 1 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(fillsgn(n, bits), true ')) = fillsgn(n, bits), # bits = 1, bits = []

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' ≠ [];% ⊦ # boolvar ' = 0 +1;