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Proof step 1 from proof square-ls-01 (conclusion)

⊦ (m * (m + 1)) / 2 < ((n + 1) * (n + 2)) / 2 ↔ m < n + 1

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: square-ls-02 seq: ⊦ (m * (m + 1)) / 2 < (n * (n + 1)) / 2 ↔ m < n subst: [m, n] → [m, n + 1]

Additional information

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