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Proof step 5 from proof square-less (previous step)

m * m < n * n, n ≤ m ⊦ n = 0

User applied rule apply lemma

And obtained the following new goals

Proof step 6:
⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1

Rule argument

spec: nat name: let seq: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1 subst: [n0, n, n1] → [n * m, n * n, m * m]

Additional information

Used simplifier rules: le: ⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%ls: ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n;%ls-06: ⊦ m < m0 ∧ n < n0 → m * n < m0 * n0;% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ ¬ n < n; (s) Used terms: n * m, n * n, m * m