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Proof step 5 from proof valid-sqns-subarray-h (previous step)

n1 < # apebs, n1 < n, n ≤ # apebs, ∀ n, apebs, v, m. n ≤ # apebs ∧ n1 ≤ n ∧ n1 ≤ # apebs → valid-sqns(v, m, subarray(apebs, 0, n), n1) = valid-sqns(v, m, apebs, n1) ⊦ valid-sqns(v, m, subarray(apebs, 0, n), n1 + 1) = valid-sqns(v, m, apebs, n1) ∪ (valid-peb(v, m, apebs[n1]) ⊃ {apebs[n1].vidhdr.sqn} ;∅)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: valid-sqns-rec seq: ⊦ n < # apebs → valid-sqns(v, m, apebs, n + 1) = valid-sqns(v, m, apebs, n) ∪ (valid-peb(v, m, apebs[n]) ⊃ {apebs[n].vidhdr.sqn} ;∅) subst: [n, apebs, m, v] → [n1, subarray(apebs, 0, n), m, v] flags: <all paths>

Additional information

Used simplifier rules: valid-sqns-rec: ⊦ n < # apebs → valid-sqns(v, m, apebs, n + 1) = valid-sqns(v, m, apebs, n) ∪ (valid-peb(v, m, apebs[n]) ⊃ {apebs[n].vidhdr.sqn} ;∅);% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ # subarray(apebs, m, n) = n; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%# apebs = n ⊦ subarray(apebs, 0, n) = apebs; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ bag0 ∪ bag3 = bag0 ∪ bag4 ↔ bag3 = bag4; (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] Used terms: n1, subarray(apebs, 0, n), m, v