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Proof step 28 from proof valid-wl-next-valid-use (previous step)

# apebs = # wlar, avhdr0.sqn = ∅.max, apebs[m] = mkpeb(aehdr, avhdr, buf, m2, boolvar), wlar[n0] = wl-entry(m3, used), apebs[n0] = mkpeb(aechdr(m3), avhdr0, buf0, m0, false), valid-sqns(v0, m1, apebs, m) = ∅, n0 < m, m < # wlar, n0 < # wlar, n0 + 1 < # wlar, valid-peb(v0, m1, mkpeb(aechdr(m3), avhdr0, buf0, m0, false)), valid-peb(v0, m1, mkpeb(aehdr, avhdr, buf, m2, boolvar)) ⊦ ∅.max = avhdr.sqn, n0 = m

User applied rule apply lemma

And obtained the following new goals

Proof step 29:
⊦ n < m0 ∧ m0 ≤ # apebs ∧ valid-peb(v, m, apebs[n]) → (∃ bag. valid-sqns(v, m, apebs, m0) = bag ++ apebs[n].vidhdr.sqn)

Rule argument

spec: sqns name: valid-sqns-extract-sqn-h seq: ⊦ n < m0 ∧ m0 ≤ # apebs ∧ valid-peb(v, m, apebs[n]) → (∃ bag. valid-sqns(v, m, apebs, m0) = bag ++ apebs[n].vidhdr.sqn) subst: [n, m0, apebs, v, m] → [n0, m, apebs, v0, m1]

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (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)% ⊦ ∅ ≠ bag ++ n; (s) Used terms: n0, m, apebs, v0, m1