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Proof step 54 from proof wlar-erase-worker-helper (previous step)

apebs0[n] = mkpeb(aehdr0, avhdr, buf, m1, false), wlar[n] = wl-entry(m0, stat0), apebs1 = apebs0[n, mkpeb(aehdr, empty, mkempbuf(LEB_SIZE), 0, false)], err = err0, vtblpnum = n, vtblpnum0 = 1, wlar0 = wlar[n, wl-entry(m0 + 1, stat0)], n < # apebs0, n < # wlar, id○(wlar, wlar1, n), vtblpnum1 ≤ UBI_ERASE_RETRIES + 1, vtblpnum1 ≤ UBI_ERASE_RETRIES, ⌊err0⌋ , ∃ stat. wlar1 = wlar[n, wl-entry(wlar[n].ec + vtblpnum1, stat)] ⊦ [wlar1[vtblpnum].ec := wlar1[vtblpnum].ec + 1; aubi_io_sync_erase#(vtblpnum; apebs; err1); if err1 = ESUCCESS then aubi_io_write_echdr#(vtblpnum, aechdr(wlar1[vtblpnum].ec); apebs; err1) else debug#("ubi: erase of PEB " + vtblpnum.toString + " failed"); vtblpnum1 := vtblpnum1 + 1] ( n < # wlar ∧ 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 symbolic execution applied rule assign right

And obtained the following new goals

Rule argument

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