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Proof step 5 from proof valid-recs-in-nonempty-sqns (previous step)

# wlar = # apebs, apebs[recs[v × n].pnum].vidhdr.leb = n, wlar[recs[v × n].pnum].status = used, apebs[recs[v × n].pnum].vidhdr.vol = v, valid-sqns(v, n, apebs, m) = ∅, recs[v × n].pnum + 1 < # apebs, m < # apebs, recs[v × n].pnum < m, recs[v × n].pnum < # apebs, valid-wl(subarray(apebs, 0, m), subarray(wlar, 0, m)), v × n ∈ recs ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
# wlar = # apebs, apebs[m0].vidhdr.leb = n, wlar[m0].status = used, apebs[m0].vidhdr.vol = v, recs[v × n] = recovery-entry(m0, m1), valid-sqns(v, n, apebs, m) = ∅, m0 < # apebs, m0 < m, m < # apebs, m0 + 1 < # apebs, valid-wl(subarray(apebs, 0, m), subarray(wlar, 0, m)), v × n ∈ recs ⊦

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 × n], m0, m1]

Additional information

Used simplifier rules: ⊦ m = rec.pnum ∧ m0 = rec.sqn ↔ rec = recovery-entry(m, m0); (elim) Used terms: recs[v × n], m0, m1