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Proof step 61 from proof refine-write (previous step)

# apebs = # wlar, wlar[n1] = wl-entry(m1, stat), avols[v][vtblpnum3] = mapped(buf2, m), vols[v][vtblpnum3] = ⌜n1⌝ , apebs[n1] = mkpeb(aehdr, avhdr, buf2, m0, false), avhdr.valid?, vtblpnum3 < # vols[v], vtblpnum3 < # avols[v], n1 < # wlar, abs(avols, apebs, vols, q), bflipset-cons(bflipset, wlar), ebm-cons(avols), is-aligned(n, EB_PAGE_SIZE), is-aligned(vtblpnum5, EB_PAGE_SIZE), ubi-io-inv(apebs), ubi-io-peb-inv(mkpeb(aehdr, avhdr, buf2, m0, false)), unique-sqns(apebs), valid-eraseq(apebs, q), valid-eraseq(wlar, q), valid-eraseq(q), valid-global-sqn(apebs, sqnum), valid-vtbl(apebs, vols, wlar, vtblpnum4), valid-wl(apebs, wlar), v ∈ vols, v ∈ avols, m ≤ vtblpnum5, n0 + n ≤ # buf, n + n0 ≤ # buf, vtblpnum5 + n ≤ LEB_SIZE, m0 ≤ vtblpnum5, ∀ v, m, n. vols[v][m] = ⌜n⌝ ∧ v ∈ vols ∧ m < # vols[v] ↔ wlar[n].status = used ∧ n < # apebs ∧ n < # wlar ∧ apebs[n].vidhdr.valid? ∧ apebs[n].vidhdr.vol = v ∧ apebs[n].vidhdr.leb = m ∧ n ≠ vtblpnum4 ⊦ n = 0, VTBL_VOLID ∈ vols, stat = used ∧ vtblpnum4 ≠ n1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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