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Proof step 262 from proof inv-recover (previous step)

# wlar1 = # wlar, # apebs0 = # wlar, avhdr0.leb = VTBL_LNUM, avhdr0.vol = VTBL_VOLID, wlar1[vtblpnum1] = wl-entry(m2, used), apebs0[vtblpnum1] = mkpeb(aehdr0, avhdr0, buf, m1, false), apebs0[m] = mkpeb(aehdr, avhdr, buf0, m0, boolvar), recs0[VTBL_VOLID × VTBL_LNUM] = recovery-entry(m, avhdr.sqn), vtblpnum1 < # wlar, nx < # wlar, bflipset-cons(bflipset, wlar), bflipset-cons(bflipset1, wlar1), indices-erasing(wlar, nx), is-encoded(to-vtbl(vols1), buf), ubi-io-inv(apebs0), ubi-io-peb-inv(mkpeb(aehdr0, avhdr0, buf, m1, false)), unique-sqns(apebs0), valid-ebatbl(apebs0, wlar1, vols1, vtblpnum1), valid-eraseq(wlar, q), valid-eraseq(apebs0, q1), valid-eraseq(wlar1, q1), valid-eraseq(q1), valid-eraseq-stronger(apebs0, q), valid-global-sqn(apebs0, sqnum), valid-global-sqn(apebs0, vtblpnum5), valid-recs(apebs0, wlar, recs0), valid-recs-sqns(apebs0, recs0), valid-wl(apebs0, wlar), VTBL_VOLID × VTBL_LNUM ∈ recs0, ∀ n. n < # wlar1 → (wlar1[n].status = erroneous ↔ apebs0[n].bad) ∧ (wlar1[n].status = free → apebs0[n].vidhdr = empty) ∧ (wlar1[n].status ≠ erroneous ∧ wlar1[n].status ≠ erasing → apebs0[n].echdr = aechdr(wlar1[n].ec)) ∧ (wlar1[n].status = used → (∃ v, m. valid-peb(v, m, apebs0[n]) ∧ apebs0[n].vidhdr.sqn = valid-sqns(v, m, apebs0).max)) ⊦ m = vtblpnum1, dups(nx), VTBL_VOLID ∈ vols1, vtblpnum1 ∈ q1, m ∈ q

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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