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Proof step 15 from proof bitsor-bi-zeroes (previous step)

¬(# bits0 = # bits2 ∧ bits3 ≠ [] ∧ bits0 ≠ [] ∧ false ' ≠ [] ∧ bits2 ≠ []), # bits0 = # bits2 ⊦ bits3 + bits0 = [], bits2int(bitsor(bits3 + bits0, false ' + bits2) + bits1) = bits2uint(bits2 + bits1), # bits3 = 1, bits3 = [], true ∈ bits3, true ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
# bits0 = # bits2 ⊦ bits3 + bits0 = [], bits2int(bitsor(bits3 + bits0, false ' + bits2) + bits1) = bits2uint(bits2 + bits1), # bits3 = 1, bits3 = [], true ∈ bits3, true ∈ bits0, bits0 ≠ [] ∧ bits2 ≠ []

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' ≠ [];