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Proof step 6 from proof bitsor-bi-ones (previous step)

bits0 = boolvar + bits2, boolvar ⊦ bits = [], bits0 = [], bits2int(bitsor(bits, bits0) + bits1) = bits2int(true ' + bits1), false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ bits = [], bits2int(bitsor(bits, true ' + bits2) + bits1) = bits2int(true ' + bits1), false ∈ bits

Rule argument

Additional information

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