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Proof step 5 from proof bitsor-ones-false (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ bits1, bits0. # bits0 < # bits → true → bits0 = [] ∨ bitsor(bits0, false ' + bits1) = bits0 ∨ false ∈ bits0 ⊦ bitsor(bits, false ' + bits1) = bits, bits = [], false ∈ bits

Proof step 5 from proof bitsor-ones-false (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information