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Proof step 3 from proof int2bytes-not-neg (previous step)

b2i(int2bytes(i).first) < 0 ∧ 0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → b2i(int2bytes(i0).first) < 0 ∧ 0 ≤ i0 → false ⊦ false

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information