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Proof step 6 from proof valid-sqns-nonempty-extract-max (previous step)

apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs, # apebs).max, n < # apebs, valid-peb(v, m, apebs[n]) ⊦ valid-sqns(v, m, apebs, # apebs) = ∅, ∃ n. n < # apebs ∧ valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs, # apebs).max

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
avhdr.sqn = valid-sqns(v, m, apebs, # apebs).max, apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n < # apebs, valid-peb(v, m, apebs[n]) ⊦ valid-sqns(v, m, apebs, # apebs) = ∅, ∃ n. n < # apebs ∧ valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs, # apebs).max

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