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Proof step 14 from proof bitsor-bi-sgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
∀ bits, bits1. # bits1 < # bits0 → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits0 = true ', bits0 = false ', # bits0 = 1, bits0 = [], bits2int(bitsor(true ', bits0)) = ~1

Rule argument

Additional information

Used simplifier rules: ⊦ bits2int(true ') = ~1;% ⊦ sgn(boolvar ') = boolvar ';% ⊦ boolvar ' ≠ [];% ⊦ # boolvar ' = 0 +1;