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Proof step 18 from proof valid-eraseq-strong-notinvols (previous step)

n < # apebs, m0 < # apebs, valid-eraseq-strong(apebs, vtbl, q0), valid-peb(v, m, apebs[n]), ∀ n, apebs, vtbl, v, m. n < # apebs ∧ valid-eraseq-strong(apebs, vtbl, q0) ∧ valid-peb(v, m, apebs[n]) ∧ (¬ v ∈ vtbl ∨ ¬ m < vtbl[v]) → valid-eraseq-strong(apebs, vtbl, q0 + eq-entry(n, v × m) ') ∨ n ∈ 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), ¬ v ∈ vtbl ∨ ¬ m < vtbl[v] ⊦ n = m0, valid-eraseq-strong(apebs, vtbl, q0 + eq-entry(n, v × m) '), n ∈ q0, m0 ∈ q0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 19:
apebs[m0] = mkpeb(aehdr, avhdr, buf, m1, boolvar), m0 < # apebs, n < # apebs, valid-eraseq-strong(apebs, vtbl, q0), valid-peb(v, m, apebs[n]), ∀ 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), ∀ n, apebs, vtbl, v, m. n < # apebs ∧ valid-eraseq-strong(apebs, vtbl, q0) ∧ valid-peb(v, m, apebs[n]) ∧ (¬ v ∈ vtbl ∨ ¬ m < vtbl[v]) → valid-eraseq-strong(apebs, vtbl, q0 + eq-entry(n, v × m) ') ∨ n ∈ q0, ¬ v ∈ vtbl ∨ ¬ m < vtbl[v] ⊦ n = m0, valid-eraseq-strong(apebs, vtbl, q0 + eq-entry(n, v × m) '), m0 ∈ q0, n ∈ 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