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Proof step 7 from proof alignup-upper-bound (previous step)

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

User applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat-div name: div-elim seq: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n subst: [m, n, m0, n0] → [m0, n, m1, n0]

Additional information

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