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Proof step 7 from proof sgn-first (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ boolvar. sgn(boolvar ' + bits0).first ↔ boolvar ⊦ sgn(boolvar ' + bits0).first ↔ boolvar

Rule argument

Additional information

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