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Proof step 9 from proof lf-valid-recs-notin-eraseq (previous step)

wlar[n0].status = used, recs[apebs[n0].vidhdr.vol × apebs[n0].vidhdr.leb] = recovery-entry(n0, n1), n0 < # apebs, valid-eraseq(wlar, q), apebs[n0].vidhdr.vol × apebs[n0].vidhdr.leb ∈ recs, n0 ∈ q, ∀ ladr, m. ladr ∈ recs ∧ recs[ladr].pnum = m ↔ wlar[m].status = used ∧ m < # apebs ∧ apebs[m].vidhdr.vol = ladr.vol ∧ apebs[m].vidhdr.leb = ladr.leb ⊦ n0 = # apebs

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
wlar[n0].status = used, recs[avhdr.vol × avhdr.leb] = recovery-entry(n0, n1), apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n0 < # apebs, valid-eraseq(wlar, q), n0 ∈ q, avhdr.vol × avhdr.leb ∈ recs, ∀ ladr, m. ladr ∈ recs ∧ recs[ladr].pnum = m ↔ wlar[m].status = used ∧ m < # apebs ∧ apebs[m].vidhdr.vol = ladr.vol ∧ apebs[m].vidhdr.leb = ladr.leb ⊦ n0 = # apebs

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[n0], 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[n0], aehdr, avhdr, buf, m0, boolvar