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Proof step 26 from proof valid-sqns-extract2-sqn-h (previous step)

apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar), apebs[m1] = mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0), n0 < m1, n0 < # apebs, m1 < # apebs, n0 + 1 < # apebs, valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), valid-peb(v, m, mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0)) ⊦ n0 = m1, ∃ bag. valid-sqns(v, m, apebs, m1) = bag ++ avhdr.sqn

User applied rule apply lemma

And obtained the following new goals

Proof step 27:
⊦ n < m0 ∧ m0 ≤ # apebs ∧ valid-peb(v, m, apebs[n]) → (∃ bag. valid-sqns(v, m, apebs, m0) = bag ++ apebs[n].vidhdr.sqn)
Proof step 28:
apebs[m1] = mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0), apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n0 + 1 < # apebs, m1 < # apebs, n0 < # apebs, n0 < m1, valid-peb(v, m, mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0)), valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), n0 < m1 ∧ m1 ≤ # apebs ∧ valid-peb(v, m, apebs[n0]) → (∃ bag. valid-sqns(v, m, apebs, m1) = bag ++ apebs[n0].vidhdr.sqn) ⊦ n0 = m1, ∃ bag. valid-sqns(v, m, apebs, m1) = bag ++ avhdr.sqn

Rule argument

spec: <toplevel> 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, m1, apebs, v, m]

Additional information

Used terms: n0, m1, apebs, v, m