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Proof step 3 from proof bitsor-first-02 (previous step)

bits0 = boolvar + bits2, bitsor(bits, bits0) = true ' + bits1, true ∈ bitsor(bits, bits0) ⊦ bits0 = [], bits = [], boolvar, bits.first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
bitsor(bits, false ' + bits2) = true ' + bits1 ⊦ bits = [], bits.first

Rule argument

Additional information

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