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Proof step 7 from proof valid-sqns-nonempty-extract-max-h (previous step)

n0 < # apebs, valid-peb(v, m, apebs[n0]), ∀ 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) ⊦ ∃ n. ¬ n0 < n ∧ valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = (valid-sqns(v, m, apebs, n0) ++ apebs[n0].vidhdr.sqn).max

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n0 < # apebs, valid-peb(v, m, apebs[n0]), ∀ 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) ⊦ ∃ n. ¬ n0 < n ∧ valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = (valid-sqns(v, m, apebs, n0) ++ avhdr.sqn).max

Rule argument

spec: apeb name: elim seq: ⊦ ( ( (aehdr = apeb.echdr ∧ avhdr = apeb.vidhdr) ∧ buf = apeb.data) ∧ m = apeb.fill) ∧ (boolvar ↔ apeb.bad) ↔ apeb = mkpeb(aehdr, avhdr, buf, m, boolvar) subst: [apeb, aehdr, avhdr, buf, m, boolvar] → [apebs[n0], aehdr, avhdr, buf, m0, boolvar]

Additional information

Used simplifier rules: ⊦ ( ((aehdr = apeb.echdr ∧ avhdr = apeb.vidhdr) ∧ buf = apeb.data) ∧ m = apeb.fill) ∧ (boolvar ↔ apeb.bad) ↔ apeb = mkpeb(aehdr, avhdr, buf, m, boolvar); (elim) Used terms: apebs[n0], aehdr, avhdr, buf, m0, boolvar