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Proof step 11 from proof valid-sqns-subseteq-h (previous step)

n0 < # apebs, n < # apebs, n0 < n, n0 + 1 < # apebs, valid-peb(v, m, apebs[n]), n0 ≤ n + 1, valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ⊦ valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ++ apebs[n].vidhdr.sqn

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n0 + 1 < # apebs, n0 < n, n < # apebs, n0 < # apebs, valid-peb(v, m, apebs[n]), n0 ≤ n + 1, valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ⊦ valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ++ 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] → [apebs[n], aehdr, avhdr, buf, m0, boolvar]

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: apebs[n], aehdr, avhdr, buf, m0, boolvar