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Proof step 4 from proof bitsor-empty (previous step)

¬(1 < # bits ∧ 1 < # bits0), bitsor(bits, bits0) = [] ⊦ bits = [], bits0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
bitsor(bits, bits0) = [] ⊦ bits = [], bits0 = [], # bits ≠ 1 ∧ # bits0 ≠ 1

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);