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Proof step 3 from proof square-ls-02 (previous step)

(m * (m + 1)) % 2 = 0 ∧ (n * (n + 1)) % 2 = 0 ⊦ m * (m + 1) < n * (n + 1) ↔ m < n

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-mult inst: name: square-ls seq: ⊦ m * (m + 1) < n * (n + 1) ↔ m < n subst: [n, m] → [n, m] flags: rewrite

Additional information

Used terms: n, m