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Proof step 4 from proof le-succ (previous step)

m * n ≤ m + 1 ⊦ m = 0, 2 ≤ m ∧ n < 2, m = 1 ∧ n ≤ 2

User applied rule apply lemma

And obtained the following new goals

Proof step 5:
n0 < m0, m0 * n ≤ m0 * m + n0 ⊦ n ≤ m

Rule argument

spec: <toplevel> name: modp seq: n0 < m0, m0 * n ≤ m0 * m + n0 ⊦ n ≤ m subst: [n0, m0, n, m] → [1, m, n, 1]

Additional information

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