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Proof step 3 from proof sgn-in-01 (previous step)

∀ boolvar. bits0 = [] ∨ (boolvar ∈ sgn(bits0) ↔ boolvar ∈ bits0) ⊦ boolvar + bits0 = [] ∨ (boolvar0 ∈ sgn(boolvar + bits0) ↔ boolvar0 ∈ boolvar + bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ boolvar. boolvar ∈ sgn(bits0) ↔ boolvar ∈ bits0 ⊦ boolvar0 ∈ sgn(boolvar ' + bits0) ↔ (boolvar0 ↔ boolvar) ∨ boolvar0 ∈ bits0, 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;