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Proof step 15 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(bits0 + bits1) = sgn((¬ boolvar0)' + bits1), boolvar0 ↔ boolvar, boolvar0 ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
bits0 = [] ∨ (∀ bits1, boolvar. sgn(bits0 + bits1) = sgn((¬ boolvar)' + bits1) ∨ boolvar ∈ bits0) ⊦ sgn(bits0 + bits1) = sgn((¬ boolvar0)' + bits1), boolvar0 ↔ (bits0 + bits1).first, bits0 + bits1 = [], boolvar0 ∈ bits0

Rule argument

Additional information

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