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Proof step 57 from proof wlar-erase-worker-helper (previous step)

wlar[n] = wl-entry(m0, stat0), apebs0[n] = mkpeb(aehdr0, avhdr, buf, m1, false), vtblpnum = n, n < # wlar, n < # apebs0, vtblpnum1 ≤ UBI_ERASE_RETRIES, vtblpnum1 ≤ UBI_ERASE_RETRIES + 1, ⌊err0⌋ ⊦ [if ubi-io-erase-pre(vtblpnum, apebs) then {fail_or_success#(; ; err); if err = ESUCCESS then apebs[vtblpnum] := mkpeb(empty, empty, mkempbuf(LEB_SIZE), 0, false) else {choose apeb with ubi-io-peb-inv(apeb) ∧ ¬ apeb.bad ∧ ¬ apeb.vidhdr.valid? in apebs[vtblpnum] := apeb or skip}} else abort; wlar1 := wlar[n, wl-entry(wlar[n].ec + vtblpnum1, stat)] [n, wlar[n, wl-entry(wlar[n].ec + vtblpnum1, stat)][n].ec:= (wlar[n, wl-entry(wlar[n].ec + vtblpnum1, stat)][n].ec + 1) ]; if err = ESUCCESS then aubi_io_write_echdr#(n, aechdr(wlar1[n].ec); apebs; err1) else debug#("ubi: erase of PEB " + n.toString + " failed"); vtblpnum1 := vtblpnum1 + 1] (id○(wlar, wlar1, n) ∧ vtblpnum1 ≠ 0 ∧ vtblpnum1 ≤ UBI_ERASE_RETRIES + 1 ∧ (∃ stat. wlar1 = wlar[n, wl-entry(wlar[n].ec + vtblpnum1, stat)])), err1 = ESUCCESS, vtblpnum1 = 0, aehdr.valid?

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

pos: right 1

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