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Proof step 40 from proof inv-format (previous step)

# apebs0 = n0 + vtblpnum0, # wlar = n0 + vtblpnum0, apebs0[vtblpnum2] = mkpeb(aehdr0, avhdr, buf, m, boolvar), err = err0, ubi-io-initial(apebs0), vtblpnum2 ≤ n0 + vtblpnum0, vtblpnum2 ≠ n0 + vtblpnum0 → ¬ boolvar, ∀ n3. n3 < vtblpnum0 → (apebs0[n3].bad → wlar[n3] = wl-entry(0, erroneous)) ∧ (¬ apebs0[n3].bad → apebs0[n3].echdr.valid? ∧ wlar[n3] = wl-entry(apebs0[n3].echdr.ec, free) ∧ apebs0[n3].vidhdr = empty), ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪if err = ESUCCESS then isbad := apebs0[vtblpnum0].bad else isbad := ?; vtblpnum := vtblpnum0; n := # apebs0; err1 := err; if err1 = ESUCCESS then if ¬ isbad then let aehdr = ?, bitflips = ? in {aubi_io_read_echdr#(vtblpnum, apebs0; aehdr; bitflips, err1); if err1 = ESUCCESS ∧ vtblpnum2 = n then vtblpnum2 := vtblpnum; if err1 = ESUCCESS ∧ aehdr.valid? then wlar[vtblpnum] := wl-entry(aehdr.ec, free)} else wlar[vtblpnum] := wl-entry(0, erroneous); vtblpnum := vtblpnum + 1⟫ ( vtblpnum ≤ # apebs0 ∧ n = # apebs0 ∧ # wlar = # apebs0 ∧ (err1 = ESUCCESS ∨ ⌊err1⌋ ) ∧ ( err1 = ESUCCESS → vtblpnum2 ≤ # apebs0 ∧ (∀ n3. n3 < vtblpnum → (apebs0[n3].bad → wlar[n3] = wl-entry(0, erroneous)) ∧ (¬ apebs0[n3].bad → apebs0[n3].echdr.valid? ∧ wlar[n3] = wl-entry(apebs0[n3].echdr.ec, free) ∧ apebs0[n3].vidhdr = empty)) ∧ (vtblpnum2 ≠ # apebs0 → ¬ apebs0[vtblpnum2].bad)) ∧ # apebs0 < n0 + vtblpnum), n0 = 0

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

pos: right 1

Additional information