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Proof step 1 from proof bitsor-sgn-in (conclusion)

⊦ bits ≠ [] ∧ bits0 ≠ [] → (boolvar ∈ bitsor(sgn(bits), bits0) ↔ boolvar ∈ bitsor(bits, bits0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ bits = [], bits0 = [], boolvar ∈ bitsor(sgn(bits), bits0) ↔ boolvar ∈ bitsor(bits, bits0)

Rule argument

Additional information