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Proof step 6 from proof sgn-append-03 (previous step)

¬(bits0 + bits1 ≠ [] ∧ ((bits0 + bits1).first ↔ boolvar)), bits0 = [] ∨ (∀ bits1, boolvar. sgn(bits0 + bits1) = sgn((¬ boolvar)' + bits1) ∨ boolvar ∈ bits0) ⊦ sgn((boolvar ' + bits0) + bits1) = sgn((¬ boolvar0)' + bits1), boolvar0 ↔ boolvar, boolvar0 ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
bits0 = [] ∨ (∀ bits1, boolvar. sgn(bits0 + bits1) = sgn((¬ boolvar)' + bits1) ∨ boolvar ∈ bits0) ⊦ sgn((boolvar ' + bits0) + bits1) = sgn((¬ boolvar0)' + bits1), boolvar0 ↔ boolvar, boolvar0 ∈ bits0, (bits0 ≠ [] ∨ bits1 ≠ []) ∧ ((bits0 + bits1).first ↔ boolvar)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ ¬ boolvar ∈ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ [] + bits = bits;