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Proof step 5 from proof bitsor-first-01 (previous step)

bits0 = boolvar0 + bits ⊦ bits0 = [], bitsor(boolvar ' + bits1, bits0).first ↔ boolvar ∨ boolvar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ bitsor(boolvar ' + bits1, boolvar0 ' + bits).first ↔ boolvar ∨ boolvar0

Rule argument

Additional information

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