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Proof step 135 from proof inv-format (previous step)

# wlar = n0 + vtblpnum0, # apebs0 = n0 + vtblpnum0, apebs0[vtblpnum2] = mkpeb(aehdr0, avhdr, buf, m, boolvar), apebs0[vtblpnum0] = mkpeb(aehdr1, avhdr0, buf0, m0, false), apebs = 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) ⊦ n0 = 0, ubi-io-read-echdr-pre(vtblpnum, apebs)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).bad ↔ boolvar; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ ubi-io-read-echdr-pre(n, apebs) ↔ n < # apebs ∧ ¬ apebs[n].bad; (s)