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Proof step 11 from proof sgn-last-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
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)

Rule argument

Additional information

Used simplifier rules: ⊦ (boolvar ' + bits).first ↔ boolvar;% ⊦ bits + boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + bits0 = bits1 + bits0 ↔ bits = bits1;