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Proof step 27 from proof valid-eraseq-stronger-next (previous step)

avhdr.sqn = avhdr0.sqn, apebs[m] = mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0), apebs[m1] = mkpeb(aehdr, avhdr, buf, m2, boolvar), valid-sqns(v, m0, apebs, m) = ∅, m1 + 1 < # apebs, m1 < m, m < # apebs, m1 < # apebs, valid-eraseq-stronger(subarray(apebs, 0, m + 1), q0), valid-eraseq-stronger(subarray(apebs, 0, m), q0), valid-peb(v, m0, mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0)), valid-peb(v, m0, mkpeb(aehdr, avhdr, buf, m2, boolvar)), ∀ m, apebs. m < # apebs ∧ valid-eraseq-stronger(subarray(apebs, 0, m), q0) → valid-eraseq-stronger(subarray(apebs, 0, m + 1), q0) ⊦ avhdr0.sqn = ∅.max, m1 ∈ q0

User applied rule apply lemma

And obtained the following new goals

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] → [m1, m, apebs, v, m0]

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: m1, m, apebs, v, m0