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Proof step 41 from proof sgn-last-04 (previous step)

bits0 = boolvar2 + bits1, boolvar2 ↔ ¬ boolvar1, bits0 + boolvar0 ' = boolvar1 ' + bits, boolvar0 ∈ bits, boolvar1 ∈ bits0, ∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ sgn(boolvar1 ' + bits) = (boolvar1 ' + bits0) + boolvar0 ', bits0 = [], ¬ (¬ boolvar0) ∈ bits0 ∧ ¬(¬ boolvar0 ↔ boolvar1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
¬ boolvar1 ↔ boolvar1, ∀ boolvar. (¬ boolvar ↔ boolvar1) ∨ (¬ boolvar) ∈ bits1 → sgn((boolvar1 ' + bits1) + boolvar ') = sgn(boolvar1 ' + bits1) + boolvar ' ⊦ sgn(boolvar1 ' + bits1 + boolvar0 ') = boolvar1 ' + (boolvar1 ' + bits1) + boolvar0 ', ¬(¬ boolvar0 ↔ boolvar1) ∧ ¬ (¬ boolvar0) ∈ bits1 ∧ ¬(¬ boolvar0 ↔ boolvar1)

Rule argument

Additional information

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