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

true, ∀ bits0. # bits0 < # bits → true → bits2uint(bits0) = 0 ∨ true ∈ bits0 ⊦ bits2uint(bits) = 0 ∨ true ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information