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Proof step 1 from proof sgn-ex-01 (conclusion)

⊦ true ∈ bits ∧ false ∈ bits → (∃ boolvar, n, boolvar0, bits0. bits = mklist(boolvar, n) + (boolvar0 + bits0) ∧ sgn(bits) = boolvar + boolvar0 + bits0 ∧ ¬(boolvar ↔ boolvar0) ∧ n ≠ 0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
true ∈ bits, false ∈ bits ⊦ ∃ boolvar, n, boolvar0, bits0. bits = mklist(boolvar, n) + boolvar0 ' + bits0 ∧ sgn(mklist(boolvar, n) + boolvar0 ' + bits0) = boolvar ' + boolvar0 ' + bits0 ∧ ¬(boolvar ↔ boolvar0) ∧ n ≠ 0

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar + bits = boolvar ' + bits;