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Proof step 2 from proof valid-eraseq-stronger-one (previous step)

⊦ valid-eraseq-stronger(apebs, eq-entry(m, lref) ') ↔ m < # apebs ∧ (lref = none → ¬ valid-peb(apebs[m])) ∧ (lref ≠ none → valid-peb(lref.vol, lref.leb, apebs[m]) ∧ apebs[m].vidhdr.sqn ≠ valid-sqns(lref.vol, lref.leb, apebs).max)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
apebs[m] = mkpeb(aehdr, avhdr, buf, m0, boolvar) ⊦ valid-eraseq-stronger(apebs, eq-entry(m, lref) ') ↔ m < # apebs ∧ (lref = none → ¬ valid-peb(apebs[m])) ∧ (lref ≠ none → valid-peb(lref.vol, lref.leb, apebs[m]) ∧ avhdr.sqn ≠ valid-sqns(lref.vol, lref.leb, 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[m], 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[m], aehdr, avhdr, buf, m0, boolvar