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Proof step 6 from proof valid-recs-in-nonempty-sqns (previous step)

# 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 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# wlar = # apebs, wlar[m0].status = used, recs[apebs[m0].vidhdr.vol × apebs[m0].vidhdr.leb] = recovery-entry(m0, m1), valid-sqns(apebs[m0].vidhdr.vol, apebs[m0].vidhdr.leb, apebs, m) = ∅, m0 < m, m < # apebs, m0 < # apebs, m0 + 1 < # apebs, valid-wl(subarray(apebs, 0, m), subarray(wlar, 0, m)), apebs[m0].vidhdr.vol × apebs[m0].vidhdr.leb ∈ recs ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)%transitivity: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification oelem[nat-set]%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic