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Proof step 17 from proof bitsor-bi-ones (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
⊦ bits3 = [], # bits3 = 1, bits2int(bitsor(bits3, true ') + bits1) = bits2int(true ' + bits1), false ∈ bits3

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ boolvar ∈ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ # bits = 0 ↔ bits = [];% ⊦ # [] = 0;