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Proof step 98 from proof refine-write-data (previous step)

pebs[m3] = mkpeb(buf1, 2 * EB_PAGE_SIZE + m2, false), apebs[m3] = mkpeb(aehdr, avhdr, buf0, m2, false), err = err1, m0 = m3, n1 = n2, aehdr.valid?, avhdr.valid?, m3 < # pebs, m3 < # apebs, n3 < n, abs(pebs, apebs), abs-peb(mkpeb(buf1, 2 * EB_PAGE_SIZE + m2, false), mkpeb(aehdr, avhdr, buf0, m2, false)), is-aligned(n3, EB_PAGE_SIZE), is-aligned(n, EB_PAGE_SIZE), is-aligned(m1, EB_PAGE_SIZE), mtd-cons(pebs), ubi-io-inv(apebs), m2 ≤ m1, (m1 + 2 * EB_PAGE_SIZE) + n3 ≤ PEB_SIZE, n2 + n ≤ # buf, m1 + n ≤ LEB_SIZE, n2 + n3 ≤ # buf, ⌊err1⌋ ⊦ ⟨m := m1; choose n2 with (n2 < n ∨ n2 = 0 ∧ n = 0) ∧ is-aligned(n2, EB_PAGE_SIZE) in if n2 ≠ 0 then apebs[m0] := mkpeb (apebs[m0].echdr, apebs[m0].vidhdr, subarray(apebs[m0].data, 0, m) + subarray(buf, n1, n2) + mkempbuf(LEB_SIZE - m - n2), m + n2, apebs[m0].bad)⟩ ( abs(pebs[m3, (pebs[m3].data:= copy(buf, n2, pebs[m3].data, m1 + 2 * EB_PAGE_SIZE, n3)).fill:= ((m1 + 2 * EB_PAGE_SIZE) + n3)], apebs) ∧ err1 = err), n3 = 0

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Rule argument

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