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Proof step 20 from proof bitsor-bi-ones (previous step)

# bits0 = # bits2 ∧ bits3 ≠ [] ∧ bits0 ≠ [] ∧ true ' ≠ [] ∧ bits2 ≠ [], # bits0 = # bits2 ⊦ bits3 + bits0 = [], bits2int((bitsor(bits3, true ') + bitsor(bits0, bits2)) + bits1) = bits2int(true ' + bits1), # bits3 = 1, bits3 = [], false ∈ bits3, false ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
# bits0 = # bits2 ⊦ bits3 + bits0 = [], bits2int((bitsor(bits3, true ') + bitsor(bits0, bits2)) + bits1) = bits2int(true ' + bits1), # bits3 = 1, bits2 = [], bits0 = [], bits3 = [], false ∈ bits3, false ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' ≠ [];