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Proof step 140 from proof inv-format (previous step)

# apebs0 = n0 + vtblpnum0, # wlar = n0 + vtblpnum0, apebs0[vtblpnum2] = mkpeb(aehdr0, avhdr, buf, m, boolvar), apebs0[n] = mkpeb(aehdr, avhdr0, buf0, m0, boolvar0), apebs0[vtblpnum0].bad, ubi-io-initial(apebs0), n ≤ vtblpnum0, 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, (boolvar0 → wlar[vtblpnum0, wl-entry(0, erroneous)][n] = wl-entry(0, erroneous)) ∧ (¬ boolvar0 → aehdr.valid? ∧ wlar[n] = wl-entry(aehdr.ec, free) ∧ avhdr0 = empty)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: apeb name: elim seq: ⊦ ( ( (aehdr = apeb.echdr ∧ avhdr = apeb.vidhdr) ∧ buf = apeb.data) ∧ m = apeb.fill) ∧ (boolvar ↔ apeb.bad) ↔ apeb = mkpeb(aehdr, avhdr, buf, m, boolvar) subst: [apeb, aehdr, avhdr, buf, m, boolvar] → [apebs0[vtblpnum0], aehdr1, avhdr1, buf1, m1, boolvar1]

Additional information

Used simplifier rules: ⊦ ( ((aehdr = apeb.echdr ∧ avhdr = apeb.vidhdr) ∧ buf = apeb.data) ∧ m = apeb.fill) ∧ (boolvar ↔ apeb.bad) ↔ apeb = mkpeb(aehdr, avhdr, buf, m, boolvar); (elim) Used terms: apebs0[vtblpnum0], aehdr1, avhdr1, buf1, m1, boolvar1