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Proof step 17 from proof alignup-le-is-aligned (previous step)

m - m / n * n = 0, m0 = n0 + m0 / n * n, n0 < n, m < n * m0 / n + n, m0 < n * m0 / n + n, m0 ≤ m ⊦ n0 = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m, m / n * n, n1]

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n → n ≠ 0; (s)%n ≠ 0 ⊦ m / n * n ≤ m; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ m * n < m1 * n + n ↔ m < m1 + 1 ∧ n ≠ 0; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%m0 ≠ 0 ⊦ n ≤ m / m0 ↔ n * m0 ≤ m; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m ≤ m; (s)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m = m + n ↔ n = 0; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m, m / n * n, n1