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Proof step 36 from proof bitsor-sgn-in (previous step)

⊦ true ∈ bits, bits = [], bits0 = [], boolvar ∈ bitsor(sgn(bits), bits0) ↔ boolvar ∈ bitsor(bits, bits0), true ∈ bits ∧ false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
⊦ bits = [], bits0 = [], boolvar ∈ bitsor(false ', bits0) ↔ boolvar ∈ bitsor(bits, bits0), true ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ true ∈ bits ∧ bits ≠ [] → sgn(bits) = false ';