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Proof step 6 from proof fill-sgn-02 (previous step)

¬(bits0 ≠ [] ∧ (bits0.first ↔ boolvar)), fillsgn(# bits0, sgn(bits0)) = bits0 ⊦ fillsgn(1 + # bits0, sgn(boolvar ' + bits0)) = boolvar ' + bits0, bits0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
fillsgn(# bits0, sgn(bits0)) = bits0 ⊦ fillsgn(1 + # bits0, sgn(boolvar ' + bits0)) = boolvar ' + bits0, bits0 = [], bits0.first ↔ boolvar

Proof step 6 from proof fill-sgn-02 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information