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Proof step 21 from proof valid-sqns-extract2-sqn-h (previous step)

apebs[m1] = mkpeb(aehdr, avhdr, buf, m0, boolvar), apebs[n1] = mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0), n1 < m1, n1 < # apebs, m1 < # apebs, n1 + 1 < # apebs, valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), valid-peb(v, m, mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0)) ⊦ m1 = n1, ∃ bag. valid-sqns(v, m, apebs, m1) ++ avhdr.sqn = bag ++ avhdr.sqn ++ avhdr0.sqn

User applied rule apply lemma

And obtained the following new goals

Proof step 22:
⊦ n < m0 ∧ m0 ≤ # apebs ∧ valid-peb(v, m, apebs[n]) → (∃ bag. valid-sqns(v, m, apebs, m0) = bag ++ apebs[n].vidhdr.sqn)
Proof step 23:
apebs[n1] = mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0), apebs[m1] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n1 + 1 < # apebs, m1 < # apebs, n1 < # apebs, n1 < m1, valid-peb(v, m, mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0)), valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), n1 < m1 ∧ m1 ≤ # apebs ∧ valid-peb(v, m, apebs[n1]) → (∃ bag. valid-sqns(v, m, apebs, m1) = bag ++ apebs[n1].vidhdr.sqn) ⊦ m1 = n1, ∃ bag. valid-sqns(v, m, apebs, m1) ++ avhdr.sqn = bag ++ avhdr.sqn ++ avhdr0.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] → [n1, m1, apebs, v, m]

Additional information

Used terms: n1, m1, apebs, v, m