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Proof step 60 from proof inv-format (previous step)

# apebs0 = n0 + vtblpnum0, # wlar = n0 + vtblpnum0, apebs0[vtblpnum0] = mkpeb(aehdr1, avhdr0, buf0, m0, false), apebs0[vtblpnum2] = mkpeb(aehdr0, avhdr, buf, m, boolvar), err = err0, err = ESUCCESS, vtblpnum0 < # apebs0, 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 ⊦ ⟪n := # apebs0; if err = ESUCCESS ∧ vtblpnum2 = n then vtblpnum2 := vtblpnum0; if err = ESUCCESS ∧ apebs0[vtblpnum0].echdr.valid? then wlar[vtblpnum0] := wl-entry(apebs0[vtblpnum0].echdr.ec, free); vtblpnum0 := vtblpnum0 + 1⟫ ( vtblpnum0 ≤ # apebs0 ∧ n = # apebs0 ∧ # wlar = # apebs0 ∧ (err = ESUCCESS ∨ ⌊err⌋ ) ∧ ( err = ESUCCESS → vtblpnum2 ≤ # apebs0 ∧ (∀ 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)) ∧ (vtblpnum2 ≠ # apebs0 → ¬ apebs0[vtblpnum2].bad)) ∧ # apebs0 < n0 + vtblpnum0), n0 = 0

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Rule argument

Additional information