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Proof step 3 from proof bitsor-ones-false (previous step)

bits0 = boolvar + bits1 ⊦ bits = [], bits0 = [], bitsor(bits, bits0) = bits, false ∈ bits, boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ bits = [], bitsor(bits, false ' + bits1) = bits, false ∈ bits

Rule argument

Additional information

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