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Proof step 25 from proof sgn-last-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
bits0 + boolvar0 ' = boolvar1 ' + bits, boolvar0 ∈ bits, boolvar1 ∈ bits0, ∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ bits0 = [], sgn(boolvar1 ' + bits) = sgn(boolvar1 ' + bits0) + boolvar0 ', bits0.first ↔ ¬ boolvar1, ¬ (¬ boolvar0) ∈ bits0 ∧ ¬(¬ boolvar0 ↔ boolvar1)

Rule argument

Additional information

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