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Proof step 6 from proof valid-recs-next-update (previous step)

# wlar = # apebs, recs[v × n] = recovery-entry(m3, m4), apebs[m] = mkpeb(aehdr, avidhdr(v, n, m0, m1, m5), buf, m2, boolvar), recs[v × n, recovery-entry(m, m0)][v0 × m7] = recovery-entry(m8, m9), m < # apebs, v × n ∈ recs, ∀ ladr, m6. ladr ∈ recs ∧ recs[ladr].pnum = m6 ↔ wlar[m6].status = used ∧ m6 < m ∧ apebs[m6].vidhdr.vol = ladr.vol ∧ apebs[m6].vidhdr.leb = ladr.leb ⊦ v0 × m7 ∈ recs[v × n, recovery-entry(m, m0)] ∧ m8 = m6 ↔ subarray(wlar[m, wl-entry(n0, used)][m3, wl-entry(n1, erasing)], 0, m + 1)[m6].status = used ∧ ¬ m < m6 ∧ subarray(apebs, 0, m + 1)[m6].vidhdr.vol = v0 ∧ subarray(apebs, 0, m + 1)[m6].vidhdr.leb = m7

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
# wlar = # apebs, recs[v × n, recovery-entry(m, m0)][v0 × m7] = recovery-entry(m8, m9), apebs[m] = mkpeb(aehdr, avidhdr(v, n, m0, m1, m5), buf, m2, boolvar), recs[v × n] = recovery-entry(m3, m4), subarray(apebs, 0, m + 1)[m6] = mkpeb(aehdr0, avhdr, buf0, m10, boolvar0), m < # apebs, v × n ∈ recs, ∀ ladr, m6. ladr ∈ recs ∧ recs[ladr].pnum = m6 ↔ wlar[m6].status = used ∧ m6 < m ∧ apebs[m6].vidhdr.vol = ladr.vol ∧ apebs[m6].vidhdr.leb = ladr.leb ⊦ v0 × m7 ∈ recs[v × n, recovery-entry(m, m0)] ∧ m8 = m6 ↔ subarray(wlar[m, wl-entry(n0, used)][m3, wl-entry(n1, erasing)], 0, m + 1)[m6].status = used ∧ ¬ m < m6 ∧ avhdr.vol = v0 ∧ avhdr.leb = m7

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] → [subarray(apebs, 0, m + 1)[m6], aehdr0, avhdr, buf0, m10, 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: subarray(apebs, 0, m + 1)[m6], aehdr0, avhdr, buf0, m10, boolvar0