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Proof step 152 from proof inv-format (previous step)

# wlar = n0 + vtblpnum0, # apebs0 = n0 + vtblpnum0, apebs0[vtblpnum2] = mkpeb(aehdr0, avhdr, buf, m, boolvar), err = err0, err1 = err, n = # apebs0, vtblpnum = vtblpnum0, 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 ⊦ ⟪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), err = ESUCCESS, n0 = 0

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ⌊err⌋ → err ≠ ESUCCESS; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ n < n; (s)