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Proof step 9 from proof aligndown-is-n (previous step)

n0 + n1 - (n0 + n1) / n * n = 0, n < n0 + n1, n0 + n1 < 2 * n ⊦ n0 + n1 = n, n = 0, n = 1

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] → [n0 + n1, (n0 + n1) / n * n, n2]

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%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%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m0 ≠ 0 → (m0 * m < m0 * n ↔ m < n); (s)% ⊦ m0 ≠ 0 → (m0 < m0 * m ↔ 1 < m); (s)% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)%n ≠ 0 ⊦ m / n < n0 → m < n0 * n; (s)% ⊦ m ≠ 0 → (m < 2 ↔ m = 1); (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n0 + n1, (n0 + n1) / n * n, n2