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Proof step 3 from proof bits2uint-ones (previous step)

true, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + 1 ∨ false ∈ bits0 ⊦ bits = [] ∨ 2 ^ n→i(# bits) = bits2uint(bits) + 1 ∨ false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + 1 ∨ false ∈ bits0 ⊦ 2 ^ n→i(# bits) = bits2uint(bits) + 1, bits = [], false ∈ bits

Proof step 3 from proof bits2uint-ones (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information