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Proof step 28 from proof bitsor-bi-ones (previous step)

¬ false ∈ bits ∧ ¬ bits0.first ∧ bits ≠ [] ∧ bits0 ≠ [] ⊦ bits = [], bits0 = [], bits2int(bits + bits1) = bits2int(true ' + bits1), false ∈ bits

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%⊦ ¬ false ∈ bits ∧ bits ≠ [] → bits2int(bits + bits0) = bits2int(true ' + bits0);