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Proof step 5 from proof sgn-first (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: sgn-base-02: ⊦ sgn(false ' + true ' + bits) = false ' + true ' + bits;%sgn-rec: ⊦ sgn(boolvar ' + boolvar ' + bits) = sgn(boolvar ' + bits);% ⊦ (boolvar ' + bits).first ↔ boolvar;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];