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Proof step 8 from proof sgn-one-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
sgn(boolvar ' + bits0) = true ', sgn(bits0) = true ' ∧ false ∈ bits0 → bits0 = [] ⊦ bits0 ≠ [] ∧ (bits0.first ↔ boolvar), ¬ false ∈ bits0 ∧ boolvar

Rule argument

Additional information

Used simplifier rules: sgn-one: ⊦ sgn(boolvar ') = boolvar ';% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;