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Proof step 6 from proof sgn-notsame (previous step)

¬(bits0 ≠ [] ∧ (bits0.first ↔ boolvar)), sgn(boolvar ' + bits0) = boolvar0 ' + boolvar0 ', ∀ boolvar1. sgn(bits0) ≠ boolvar1 ' + boolvar1 ' ∨ ¬ boolvar1 ∈ sgn(bits0) ⊦ ¬ boolvar0 ∈ bits0 ∧ ¬(boolvar0 ↔ boolvar)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: sgn-one: ⊦ sgn(boolvar ') = boolvar ';%sgn-in-01: ⊦ bits ≠ [] → (boolvar ∈ sgn(bits) ↔ boolvar ∈ bits);% ⊦ boolvar ' ≠ [];% ⊦ boolvar ' = boolvar0 ' + bits ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;