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Proof step 2 from proof valid-sqns-nonempty-extract-max-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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