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Proof step 1 from proof square-ls-02 (conclusion)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ m % 2 = 0 ∧ n % 2 = 0 → (m / 2 < n / 2 ↔ m < n)
Proof step 3:
(m * (m + 1)) % 2 = 0 ∧ (n * (n + 1)) % 2 = 0 ⊦ m * (m + 1) < n * (n + 1) ↔ m < n

Rule argument

spec: nat-div inst: name: ls-even seq: ⊦ m % 2 = 0 ∧ n % 2 = 0 → (m / 2 < n / 2 ↔ m < n) subst: [n, m] → [n * (n + 1), m * (m + 1)] flags: rewrite, current

Additional information

Used simplifier rules: square-even-02: ⊦ (m * m + m) % 2 = 0;%rsucc-02: ⊦ m * (n + 1) = m * n + m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: n * (n + 1), m * (m + 1)