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Proof step 16 from proof bitsor-bi-zeroes (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 17:
# bits0 = # bits2 ⊦ bits0 ≠ [], bits3 + bits0 = [], bits2int(bitsor(bits3 + bits0, false ' + bits2) + bits1) = bits2uint(bits2 + bits1), # bits3 = 1, bits3 = [], true ∈ bits3, true ∈ bits0, bits0 ≠ [] ∧ bits2 ≠ []

Rule argument

fma: bits0 ≠ []

Additional information

Used simplifier rules: ⊦ # [] = 0;% ⊦ # bits = 0 ↔ bits = []; Used formulas: bits0 ≠ []