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Proof step 11 from proof bitsor-ones-false (previous step)

¬(false ' + bits1 ≠ [] ∧ bits = false '), ∀ bits1, bits0. # bits0 < # bits → true → bits0 = [] ∨ bitsor(bits0, false ' + bits1) = bits0 ∨ false ∈ bits0 ⊦ bitsor(bits, false ' + bits1) = bits, bits = [], false ∈ bits, # bits ≠ 1 ∧ bits1 ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
∀ bits1, bits0. # bits0 < # bits → true → bits0 = [] ∨ bitsor(bits0, false ' + bits1) = bits0 ∨ false ∈ bits0 ⊦ bitsor(bits, false ' + bits1) = bits, bits = [], bits = false ', false ∈ bits, # bits ≠ 1 ∧ bits1 ≠ []

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];