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Proof step 5 from proof valid-recs-next-samerecs (previous step)

# wlar = # apebs, m < # apebs, ∀ ladr, m1. ladr ∈ recs ∧ recs[ladr].pnum = m1 ↔ wlar[m1].status = used ∧ m1 < m ∧ apebs[m1].vidhdr.vol = ladr.vol ∧ apebs[m1].vidhdr.leb = ladr.leb ⊦ v × m2 ∈ recs ∧ recs[v × m2].pnum = m1 ↔ subarray(wlar[m, wl-entry(m0, stat)], 0, m + 1)[m1].status = used ∧ ¬ m < m1 ∧ subarray(apebs, 0, m + 1)[m1].vidhdr.vol = v ∧ subarray(apebs, 0, m + 1)[m1].vidhdr.leb = m2, stat = used

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
# wlar = # apebs, recs[v × m2] = recovery-entry(m3, m4), m < # apebs, ∀ ladr, m1. ladr ∈ recs ∧ recs[ladr].pnum = m1 ↔ wlar[m1].status = used ∧ m1 < m ∧ apebs[m1].vidhdr.vol = ladr.vol ∧ apebs[m1].vidhdr.leb = ladr.leb ⊦ v × m2 ∈ recs ∧ m3 = m1 ↔ subarray(wlar[m, wl-entry(m0, stat)], 0, m + 1)[m1].status = used ∧ ¬ m < m1 ∧ subarray(apebs, 0, m + 1)[m1].vidhdr.vol = v ∧ subarray(apebs, 0, m + 1)[m1].vidhdr.leb = m2, stat = used

Rule argument

spec: recovery-entry name: elim seq: ⊦ m = rec.pnum ∧ m0 = rec.sqn ↔ rec = recovery-entry(m, m0) subst: [rec, m, m0] → [recs[v × m2], m3, m4]

Additional information

Used simplifier rules: ⊦ m = rec.pnum ∧ m0 = rec.sqn ↔ rec = recovery-entry(m, m0); (elim) Used terms: recs[v × m2], m3, m4