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

∀ boolvar. ¬(sgn(bits0) = boolvar ' + boolvar ' ∧ boolvar ∈ sgn(bits0)) ⊦ ¬(sgn(boolvar + bits0) = boolvar0 ' + boolvar0 ' ∧ boolvar0 ∈ sgn(boolvar + bits0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: sgn-in-02: ⊦ boolvar ∈ sgn(boolvar0 ' + bits) ↔ boolvar ∈ boolvar0 ' + bits;% ⊦ bits + boolvar ' = bits0 → boolvar ∈ bits0; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar + bits = boolvar ' + bits;