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

n < # apebs, n ≤ n0, n0 ≤ # apebs, n0 ≤ n + 1, ∀ apebs, n0, v, m. n ≤ # apebs ∧ n0 ≤ n ∧ n0 ≤ # apebs → valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ⊦ valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ∪ (valid-peb(v, m, apebs[n]) ⊃ {apebs[n].vidhdr.sqn} ;∅)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat name: anti seq: ⊦ m ≤ n → (n ≤ m ↔ m = n) subst: [m, n] → [n0, n] 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 ≤ m; (s)% ⊦ m < n → m ≤ n; (s)% ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ bag1 ⊆ bag1 ∪ bag2; (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]% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m + 1 ≠ 0; (s)% ⊦ m = m + n ↔ n = 0; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ bag0 ∪ bag3 ⊆ bag0 ∪ bag4 ↔ bag3 ⊆ bag4; (s)% ⊦ bag0 ⊆ bag0; (s) Used terms: n0, n