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Proof step 19 from proof valid-sqns-nonempty-extract-max-h (previous step)

avhdr0.sqn = valid-sqns(v, m, apebs, n0).max, apebs[n] = mkpeb(aehdr0, avhdr0, buf0, m1, boolvar0), apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n < # apebs, n < n0, avhdr.sqn < avhdr0.sqn, n0 < # apebs, n + 1 < # apebs, valid-peb(v, m, mkpeb(aehdr0, avhdr0, buf0, m1, boolvar0)), valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), ∀ 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) ⊦ avhdr.sqn = avhdr0.sqn, valid-sqns(v, m, apebs, n0) = ∅, ∃ n. ¬ n0 < n ∧ valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = avhdr0.sqn

Heuristic quantifier applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: n

Additional information

Used simplifier rules: ⊦ m < n → ¬ n < m; (s)%⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s) Used terms: n