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Proof step 7 from proof valid-eraseq-stronger-next-append-lref (previous step)

apebs[m] = mkpeb(aehdr, avhdr, buf, m0, boolvar), m < # apebs ⊦ valid-peb(v, n, subarray(apebs, 0, m + 1)[m]) ∧ subarray(apebs, 0, m + 1)[m].vidhdr.sqn ≠ (valid-sqns(v, n, apebs, m) ∪ (valid-peb(v, n, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ⊃ {avhdr.sqn} ;∅)).max ↔ valid-peb(v, n, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ∧ valid-sqns(v, n, apebs, m) ≠ ∅ ∧ avhdr.sqn < valid-sqns(v, n, apebs, m).max

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
apebs[m] = mkpeb(aehdr, avhdr, buf, m0, boolvar), subarray(apebs, 0, m + 1)[m] = mkpeb(aehdr0, avhdr0, buf0, m1, boolvar0), m < # apebs ⊦ valid-peb(v, n, subarray(apebs, 0, m + 1)[m]) ∧ avhdr0.sqn ≠ (valid-sqns(v, n, apebs, m) ∪ (valid-peb(v, n, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ⊃ {avhdr.sqn} ;∅)).max ↔ valid-peb(v, n, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ∧ valid-sqns(v, n, apebs, m) ≠ ∅ ∧ avhdr.sqn < valid-sqns(v, n, apebs, m).max

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] → [subarray(apebs, 0, m + 1)[m], aehdr0, avhdr0, buf0, m1, boolvar0]

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: subarray(apebs, 0, m + 1)[m], aehdr0, avhdr0, buf0, m1, boolvar0