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Proof step 3 from proof uint2bits-first (previous step)

0 ≤ i ∧ uint2bits(i).first, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ uint2bits(i0).first → false ⊦ false

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information