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Proof step 7 from proof in-valid-eraseqp-valid-lref (previous step)

m0 < # apebs, valid-eraseq(apebs, q0), valid-peb(v, m, apebs[n]), valid-peb(apebs[m0]) → lref0 = apebs[m0].vidhdr.vol × apebs[m0].vidhdr.leb, ∀ apebs, v, m, n, lref. valid-eraseq(apebs, q0) ∧ valid-peb(v, m, apebs[n]) ∧ eq-entry(n, lref) ∈ q0 → lref = v × m, eq-entry(n, lref) = eq-entry(m0, lref0) ∨ eq-entry(n, lref) ∈ q0 ⊦ lref = v × m

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(apebs, q0), valid-peb(v, m, apebs[n]), valid-peb(apebs[m0]) → lref0 = avhdr.vol × avhdr.leb, ∀ apebs, v, m, n, lref. valid-eraseq(apebs, q0) ∧ valid-peb(v, m, apebs[n]) ∧ eq-entry(n, lref) ∈ q0 → lref = v × m, eq-entry(n, lref) = eq-entry(m0, lref0) ∨ eq-entry(n, lref) ∈ q0 ⊦ lref = v × m

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