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Proof step 3 from proof fill-first (previous step)

bits = boolvar + bits0 ⊦ bits = [], fillsgn(n, bits).first ↔ boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ fillsgn(n, boolvar ' + bits0).first ↔ boolvar

Rule argument

Additional information

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