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Proof step 2 from proof square-nle-01-01 (previous step)

((n + 1) * (n + 2)) / 2 ≤ n ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ (n * (n + 1)) / 2 ≤ n ↔ n < 2

Rule argument

spec: <toplevel> name: square-nle-01 seq: ⊦ (n * (n + 1)) / 2 ≤ n ↔ n < 2 subst: [n] → [n + 1]

Additional information

Used simplifier rules: two-rec: ⊦ (m + 2) / 2 = m / 2 + 1;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m * (n + 1) = m * n + m; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n; (s)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m < n → m ≤ n; (s)% ⊦ m < n → n ≠ 0; (s) Used terms: n + 1