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Proof step 6 from proof lf-valid-recs-notin-eraseq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
recs[v × m] = recovery-entry(n0, n1), valid-eraseq(wlar, q), v × m ∈ 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

Rule argument

Additional information

Used simplifier rules: ⊦ recovery-entry(m, m0).pnum = m; (s)% ⊦ ¬ n < n; (s)