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Proof step 6 from proof sgn-zeros (previous step)

¬(bits0 + false ' + bits1 ≠ [] ∧ ((bits0 + false ' + bits1).first ↔ false)), ∀ bits. sgn(bits0 + false ' + bits) = sgn(false ' + bits) ⊦ sgn((false ' + bits0) + false ' + bits1) = sgn(false ' + bits1), true ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
(bits0 + false ').first, ∀ bits. sgn(bits0 + false ' + bits) = sgn(false ' + bits) ⊦ sgn((false ' + bits0) + false ' + bits1) = sgn(false ' + bits1), true ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + boolvar ' + bits0).first ↔ (bits + boolvar ').first;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];