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Proof step 5 from proof bitsor-zeroes-true (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Proof step 5 from proof bitsor-zeroes-true (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information