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Proof step 15 from proof sgn-one-01 (previous step)

bits0 ≠ [] ∧ (bits0.first ↔ boolvar), sgn(bits0) = true ', sgn(bits0) = true ' ∧ false ∈ bits0 → bits0 = [] ⊦ ¬ false ∈ bits0 ∧ boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: sgn-true: ⊦ ¬ false ∈ bits ∧ bits ≠ [] → sgn(bits) = true ';% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;