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Proof step 10 from proof sgn-notsame (previous step)

bits0 ≠ [] ∧ (bits0.first ↔ boolvar), sgn(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

Proof step 11:
sgn(bits0) = boolvar0 ' + boolvar0 ', boolvar0 ∈ bits0, ∀ boolvar. ¬ boolvar ∈ bits0 ∨ sgn(bits0) ≠ boolvar ' + boolvar ' ⊦ bits0 = []

Rule argument

Additional information

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