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Proof step 15 from proof valid-recs-in-nonempty-sqns (previous step)

# wlar = # apebs, avhdr.sqn = ∅.max, wlar[m0] = wl-entry(m3, used), recs[v × m4] = recovery-entry(m0, m1), apebs[m0] = mkpeb(aechdr(m3), avhdr, buf, m2, false), valid-sqns(v, m4, apebs, m) = ∅, m0 + 1 < # apebs, m0 < # apebs, m < # apebs, m0 < m, valid-peb(v, m4, mkpeb(aechdr(m3), avhdr, buf, m2, false)), v × m4 ∈ recs ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 16:
n0 < n1, n1 ≤ # apebs ⊦ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ∧ apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar) → avhdr.sqn ∈ valid-sqns(v, m, apebs, n1)

Rule argument

spec: sqns name: valid-peb-in-sqnums-h seq: n0 < n1, n1 ≤ # apebs ⊦ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ∧ apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar) → avhdr.sqn ∈ valid-sqns(v, m, apebs, n1) subst: [n0, n1, apebs, v, m, aehdr, avhdr, buf, m0, boolvar] → [m0, m, apebs, v, m4, aechdr(m3), avhdr, buf, m2, false]

Additional information

Used simplifier rules: ⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)% ⊦ n ∈ bag → (bag = ∅ ↔ false); (s)% ⊦ m < n → m ≤ n; (s) Used terms: m0, m, apebs, v, m4, aechdr(m3), avhdr, buf, m2, false