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Proof step 8 from proof sgn-notsame-01 (previous step)

bits0 = boolvar1 + bits, sgn(boolvar ' + bits0) = boolvar0 ' + boolvar0 ' + bits1, ∀ boolvar, bits1. sgn(bits0) ≠ boolvar ' + boolvar ' + bits1 ⊦ boolvar1 ↔ boolvar, bits0 = [], ¬ boolvar0 ∈ bits0 ∧ ¬(boolvar0 ↔ boolvar)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
sgn(boolvar ' + boolvar1 ' + bits) = boolvar0 ' + boolvar0 ' + bits1, ∀ boolvar2, bits0. sgn(boolvar1 ' + bits) ≠ boolvar2 ' + boolvar2 ' + bits0 ⊦ boolvar1 ↔ boolvar, ¬(boolvar0 ↔ boolvar) ∧ ¬ boolvar0 ∈ bits ∧ ¬(boolvar0 ↔ boolvar1)

Rule argument

Additional information

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