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Proof step 3 from proof sgn-zero (previous step)

bits0 = [] ∨ sgn(bits0) = false ' ∨ true ∈ bits0 ⊦ boolvar + bits0 = [] ∨ sgn(boolvar + bits0) = false ' ∨ true ∈ boolvar + bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
sgn(bits0) = false ' ⊦ sgn(false ' + bits0) = false ', bits0 = [], true ∈ bits0

Rule argument

Additional information

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