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Proof step 23 from proof valid-sqns-write-h (previous step)

avhdr.sqn = avhdr0.sqn, apebs[n] = mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0), apebs[m1] = mkpeb(aehdr, avhdr, buf, m0, boolvar), valid-sqns(v, m, apebs[n, apeb], m1) = (valid-sqns(v, m, apebs, m1) -- avhdr0.sqn) ∪ (valid-peb(v, m, apeb) ⊃ {apeb.vidhdr.sqn} ;∅), n < # apebs, n < m1, m1 < # apebs, n + 1 < # apebs, valid-peb(v, m, mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0)), valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ⊦ valid-sqns(v, m, apebs[n, apeb], m1) ++ avhdr0.sqn = valid-sqns(v, m, apebs, m1) ∪ (valid-peb(v, m, apeb) ⊃ {apeb.vidhdr.sqn} ;∅), m1 = n

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 24:
avhdr.sqn = avhdr0.sqn, apebs[m1] = mkpeb(aehdr, avhdr, buf, m0, boolvar), apebs[n] = mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0), valid-sqns(v, m, apebs[n, apeb], m1) = (valid-sqns(v, m, apebs, m1) -- avhdr0.sqn) ∪ (valid-peb(v, m, apeb) ⊃ {apeb.vidhdr.sqn} ;∅), n + 1 < # apebs, m1 < # apebs, n < m1, n < # apebs, valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), valid-peb(v, m, mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0)) ⊦ valid-sqns(v, m, apebs[n, apeb], m1) ++ avhdr0.sqn = valid-sqns(v, m, apebs, m1) ∪ ∅, m1 = n, valid-peb(v, m, apeb)

Rule argument

pos: right 1

Additional information

Used simplifier rules: valid-peb-in-sqnums-h: n0 < n1, n1 ≤ # apebs ⊦ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ∧ apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar) → avhdr.sqn ∈ valid-sqns(v, m, apebs, n1);%⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)% ⊦ bag ∪ {n} = bag ++ n; (s)%union-ass: ⊦ bag ∪ bag3 ∪ bag4 = (bag ∪ bag3) ∪ bag4; from specification bag-union[nat-bag]%union-com: ⊦ bag ∪ bag2 = bag2 ∪ bag; from specification bag-union[nat-bag]% ⊦ bag ++ n ++ n0 = bag2 ++ n ↔ bag ++ n0 = bag2; (ss)% ⊦ (bag -- n) ++ n = (n ∈ bag ⊃ bag;bag ++ n); (s)% ⊦ m < n → m ≤ n; (s)