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Proof step 22 from proof and-sgn (previous step)

bits ≠ [] ∧ bits0 = false ', ∀ bits1, bits. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ sgn(bitsand(sgn(bits), bits0)) = sgn(false '), bits0 = [], bits = [], bits0 = true ', bits ≠ bits.first ' ∧ bits0 ≠ bits0.first '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
∀ bits1, bits. # bits1 < # false ' → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ bits = [], sgn(bitsand(sgn(bits), false ')) = false '

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar '.first ↔ boolvar;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ' ≠ [];% ⊦ sgn(boolvar ') = boolvar ';