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Proof step 6 from proof bitsor-sgn-in (previous step)

true ∈ bits ∧ false ∈ bits, true ∈ bits ∧ false ∈ bits → bits = mklist(boolvar0, n) + boolvar1 ' + bits1 ∧ sgn(mklist(boolvar0, n) + boolvar1 ' + bits1) = boolvar0 ' + boolvar1 ' + bits1 ∧ ¬(boolvar0 ↔ boolvar1) ∧ n ≠ 0 ⊦ bits = [], bits0 = [], boolvar ∈ bitsor(sgn(bits), bits0) ↔ boolvar ∈ bitsor(bits, bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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