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Proof step 21 from proof sgn-last-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
bits0 + boolvar0 ' = boolvar1 ' + bits, ∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ sgn(boolvar1 ' + bits) = sgn(boolvar1 ' + bits0) + boolvar0 ', bits0 ≠ [] ∧ (bits0.first ↔ ¬ boolvar1), ¬(boolvar1 ↔ boolvar0) ∧ ¬ boolvar1 ∈ bits0, ¬ boolvar0 ∈ bits ∧ ¬(boolvar0 ↔ boolvar1), ¬ (¬ boolvar0) ∈ bits0 ∧ ¬(¬ boolvar0 ↔ boolvar1)

Rule argument

Additional information

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