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Proof step 65 from proof inv-format (previous step)

# apebs0 = n0 + vtblpnum0, # wlar = n0 + vtblpnum0, apebs0[vtblpnum0] = mkpeb(aehdr1, avhdr0, buf0, m0, false), apebs0[n0 + vtblpnum0] = mkpeb(aehdr0, avhdr, buf, m, boolvar), apebs0[vtblpnum0].echdr.valid?, vtblpnum0 < # apebs0, ubi-io-initial(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) ⊦ ⟪wlar[vtblpnum0] := wl-entry(apebs0[vtblpnum0].echdr.ec, free); vtblpnum := vtblpnum0 + 1⟫ ( vtblpnum ≤ # apebs0 ∧ # wlar = # 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)) ∧ (vtblpnum0 ≠ # apebs0 → ¬ apebs0[vtblpnum0].bad) ∧ # apebs0 < n0 + vtblpnum), n0 = 0

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: nat-sum-leq: ⊦ m = n0 + n1 → (m ≤ n1 ↔ n0 = 0);% ⊦ m < n ↔ ¬ n ≤ m; (ss)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic