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Proof step 6 from proof square-ls-04 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat inst: name: fle seq: ⊦ m < n ∧ n ≤ n0 → m < n0 subst: [m, n, n0] → [m * m + m0, n * n, m * m] flags: rewrite

Additional information

Used simplifier rules: square-le-02: ⊦ m * m ≤ n * n ↔ m ≤ n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%l-02: ⊦ ¬ m + 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 * m + m0, n * n, m * m