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Proof step 5 from proof bitsor-bi-sgn (previous step)

¬(1 < # bits ∧ 1 < # bits0), ∀ 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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ 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)), # bits ≠ 1 ∧ # bits0 ≠ 1

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);