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Proof step 3 from proof sorted-2 (previous step)

m - n = 2, ar[n0 + 1] < ar[n0], n0 + 1 < m, n0 + 1 < # ar, m ≤ # ar, n ≤ n0 ⊦ ar[n + 1] < ar[n]

User applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [n, m, n0] → [n, m, n1]

Additional information

Used simplifier rules: ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%n < n0 ⊦ m + (* n) < m0 + (* n0) ↔ m < m0 + (n0 - n); (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m +1 = m + 1; (s)% ⊦ m + m0 < n → ¬ n + n0 < m; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, m, n1