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Proof step 13 from proof sgn-last-diff (previous step)

¬(boolvar ' ≠ [] ∧ (boolvar '.first ↔ ¬ boolvar0)), (¬ boolvar) ∈ bits0 ⊦ sgn(boolvar0 ' + boolvar ') = boolvar0 ' + boolvar ', (¬ boolvar0) ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
(¬ boolvar) ∈ bits0 ⊦ sgn(boolvar0 ' + boolvar ') = boolvar0 ' + boolvar ', boolvar ↔ ¬ boolvar0, (¬ boolvar0) ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar '.first ↔ boolvar;% ⊦ boolvar ' ≠ [];