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Proof step 4 from proof bitsor-bi-ones (previous step)

¬(¬ false ∈ bits ∧ ¬ bits0.first ∧ bits ≠ [] ∧ bits0 ≠ []) ⊦ 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 5:
bits0.first ⊦ bits = [], bits0 = [], bits2int(bitsor(bits, bits0) + bits1) = bits2int(true ' + bits1), false ∈ bits

Proof step 4 from proof bitsor-bi-ones (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information