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Proof step 1 from proof valid-sqns-nonempty-extract-max-h (conclusion)

⊦ n ≤ # apebs ∧ valid-sqns(v, m, apebs, n) ≠ ∅ → (∃ n0. n0 < n ∧ valid-peb(v, m, apebs[n0]) ∧ apebs[n0].vidhdr.sqn = valid-sqns(v, m, apebs, n).max)

System applied rule normalize

And obtained the following new goals

Rule argument

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