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Proof step 20 from proof bitsor-bi-sgn (previous step)

bits ≠ [] ∧ bits0 = false ', ∀ bits, bits1. # bits1 < # bits0 → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits = [], bits0 = [], bits2int(bitsor(sgn(bits), bits0)) = bits2int(bits), bits0 = true ', # bits ≠ 1 ∧ # bits0 ≠ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
∀ bits, bits1. # bits1 < # false ' → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits2int(bitsor(sgn(bits), false ')) = bits2int(bits), bits = []

Rule argument

Additional information

Used simplifier rules: ⊦ # boolvar ' = 0 +1;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ' ≠ [];