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Proof step 11 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(bitsor(bits, bits0)), bits0 = true ', # bits ≠ 1 ∧ # bits0 ≠ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# 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(bitsor(bits, bits0)), bits0 = true ', bits0 = false ', # bits0 = 1

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 1 ∧ bits ≠ true ' → bits = false '; (forward)% ⊦ # bits = 1 ∧ bits ≠ false ' → bits = true '; (forward)