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Proof step 13 from proof bitsor-ones-false (previous step)

# bits ≠ 1, ∀ 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 ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
∀ bits1, bits0. # bits0 < # bits → true → bits0 = [] ∨ bitsor(bits0, false ' + bits1) = bits0 ∨ false ∈ bits0 ⊦ # bits = 1, bits = false ', bits = [], bitsor(bits, false ') = bits, false ∈ bits

Rule argument

Additional information

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