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Proof step 7 from proof valid-sqns-extract-sqn-h (previous step)

n < # apebs, m1 < # apebs, valid-peb(v, m, apebs[n]), n ≤ m1, n < m1 → valid-sqns(v, m, apebs, m1) = bag ++ apebs[n].vidhdr.sqn ⊦ ∃ bag. valid-sqns(v, m, apebs, m1) ∪ (valid-peb(v, m, apebs[m1]) ⊃ {apebs[m1].vidhdr.sqn} ;∅) = bag ++ apebs[n].vidhdr.sqn

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), m1 < # apebs, n < # apebs, valid-peb(v, m, apebs[n]), n ≤ m1, n < m1 → valid-sqns(v, m, apebs, m1) = bag ++ avhdr.sqn ⊦ ∃ bag. valid-sqns(v, m, apebs, m1) ∪ (valid-peb(v, m, apebs[m1]) ⊃ {apebs[m1].vidhdr.sqn} ;∅) = bag ++ 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