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Proof step 18 from proof sgn-last-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: sgn-base-04: ⊦ ¬ boolvar ∈ bits → sgn(boolvar ' + bits) = boolvar ' + bits;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ bits + boolvar ' = bits0 → boolvar ∈ bits0; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;