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Proof step 3 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 structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
n0 +1 ≤ # apebs, ∀ apebs, v, m. n0 ≤ # apebs → valid-sqns(v, m, apebs, n0) = ∅ ∨ (∃ n1. n1 < n0 ∧ valid-peb(v, m, apebs[n1]) ∧ apebs[n1].vidhdr.sqn = valid-sqns(v, m, apebs, n0).max) ⊦ valid-sqns(v, m, apebs, n0 +1) = ∅, ∃ n1. n1 < n0 +1 ∧ valid-peb(v, m, apebs[n1]) ∧ apebs[n1].vidhdr.sqn = valid-sqns(v, m, apebs, n0 +1).max

Rule argument

term: n

Additional information

Used simplifier rules: valid-sqns-base: ⊦ valid-sqns(v, m, apebs, 0) = ∅;%nat freely generated by 0, +1;;