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Proof step 1 from proof sqr-08 (conclusion)

⊦ m * m ≤ 2 ↔ m ≤ 1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: le-04 seq: ⊦ m ≤ n → (n * n0 ≤ m ↔ n0 = 0 ∨ n0 = 1 ∧ m = n ∨ n = 0) subst: [m, n0, n] → [2, m, m] flags: <all paths>

Additional information

Used simplifier rules: sqzero: ⊦ m * m = 0 ↔ m = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%sqr-03: ⊦ m + 1 < m * m ↔ 1 < m;%ls-04: ⊦ m0 ≠ 0 → (m0 < m0 * m ↔ 1 < m);% ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); (s)%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat% ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic% ⊦ m +1 = m + 1; (s)% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≠ 1 → (m < 2 ↔ m = 0); (s) Used terms: 2, m, m