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Proof step 7 from proof valid-eraseq-stronger-prev-append-lref (previous step)

apebs[m] = mkpeb(aehdr, avhdr, buf, m1, boolvar), m0 < # apebs, valid-sqns(v, n, apebs, m).max < avhdr.sqn, m0 < m, m < # apebs, m0 + 1 < # apebs, valid-peb(v, n, mkpeb(aehdr, avhdr, buf, m1, boolvar)), valid-peb(v, n, apebs[m0]) ⊦ valid-peb(v, n, subarray(apebs, 0, m + 1)[m0]) ∧ subarray(apebs, 0, m + 1)[m0].vidhdr.sqn ≠ avhdr.sqn

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
apebs[m] = mkpeb(aehdr, avhdr, buf, m1, boolvar), subarray(apebs, 0, m + 1)[m0] = mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0), m0 + 1 < # apebs, m < # apebs, m0 < m, valid-sqns(v, n, apebs, m).max < avhdr.sqn, m0 < # apebs, valid-peb(v, n, apebs[m0]), valid-peb(v, n, mkpeb(aehdr, avhdr, buf, m1, boolvar)) ⊦ valid-peb(v, n, subarray(apebs, 0, m + 1)[m0]) ∧ avhdr0.sqn ≠ avhdr.sqn

Rule argument

spec: apeb name: elim seq: ⊦ ( ( (aehdr = apeb.echdr ∧ avhdr = apeb.vidhdr) ∧ buf = apeb.data) ∧ m = apeb.fill) ∧ (boolvar ↔ apeb.bad) ↔ apeb = mkpeb(aehdr, avhdr, buf, m, boolvar) subst: [apeb, aehdr, avhdr, buf, m, boolvar] → [subarray(apebs, 0, m + 1)[m0], aehdr0, avhdr0, buf0, m2, boolvar0]

Additional information

Used simplifier rules: ⊦ ( ((aehdr = apeb.echdr ∧ avhdr = apeb.vidhdr) ∧ buf = apeb.data) ∧ m = apeb.fill) ∧ (boolvar ↔ apeb.bad) ↔ apeb = mkpeb(aehdr, avhdr, buf, m, boolvar); (elim) Used terms: subarray(apebs, 0, m + 1)[m0], aehdr0, avhdr0, buf0, m2, boolvar0