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Proof step 7 from proof valid-eraseq-strong-imp-q (previous step)

m0 < # apebs, valid-eraseq-strong(apebs, vtbl, q0), (∃ apebs, vtbl. valid-eraseq-strong(apebs, vtbl, q0)) → valid-eraseq(q0), ∀ v, m. (valid-peb(v, m, apebs[m0]) ↔ lref = v × m) ∧ (lref = v × m ∧ v ∈ vtbl ∧ m < vtbl[v] → apebs[m0].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max) ⊦ valid-eraseq(eq-entry(m0, lref) ' + q0), m0 ∈ q0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
apebs[m0] = mkpeb(aehdr, avhdr, buf, m1, boolvar), m0 < # apebs, valid-eraseq-strong(apebs, vtbl, q0), (∃ apebs, vtbl. valid-eraseq-strong(apebs, vtbl, q0)) → valid-eraseq(q0), ∀ v, m. (valid-peb(v, m, apebs[m0]) ↔ lref = v × m) ∧ (lref = v × m ∧ v ∈ vtbl ∧ m < vtbl[v] → avhdr.sqn ≠ valid-sqns(v, m, apebs).max) ⊦ valid-eraseq(eq-entry(m0, lref) ' + q0), m0 ∈ q0

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] → [apebs[m0], aehdr, avhdr, buf, m1, boolvar]

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: apebs[m0], aehdr, avhdr, buf, m1, boolvar