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Proof step 11 from proof square-ls-04 (previous step)

n * n < (m + 1) * (m + 1), m * m + m0 < (m + 1) * (m + 1), m < n, m0 ≤ 2 * m, n * n ≤ m * m + m0 ⊦ n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: m + 1, n