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Proof step 14 from proof sgn-last-04 (previous step)

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