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Proof step 9 from proof valid-eraseqp-append (previous step)

apebs[n] = mkpeb(aehdr, avhdr, buf, m, boolvar), m0 < # apebs, valid-eraseq(apebs, q0), valid-peb(apebs[m0]) → lref0 = apebs[m0].vidhdr.vol × apebs[m0].vidhdr.leb, ∀ apebs, n, aehdr, avhdr, buf, m, boolvar, lref. apebs[n] = mkpeb(aehdr, avhdr, buf, m, boolvar) ∧ valid-eraseq(apebs, q0) → ( valid-eraseq(apebs, q0 + eq-entry(n, lref) ') ↔ n < # apebs ∧ (valid-peb(mkpeb(aehdr, avhdr, buf, m, boolvar)) → lref = avhdr.vol × avhdr.leb)) ⊦ valid-eraseq(apebs, q0 + eq-entry(n, lref) ') ↔ n < # apebs ∧ (valid-peb(mkpeb(aehdr, avhdr, buf, m, boolvar)) → lref = avhdr.vol × avhdr.leb)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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[m0], aehdr0, avhdr0, buf0, m1, 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: apebs[m0], aehdr0, avhdr0, buf0, m1, boolvar0