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Proof step 10 from proof is-aligned-is-aligned-mod (previous step)

n0 % n = 0, m0 * n0 + n2 = m2 * n, m1 * n + n2 < n0, n2 < n ⊦ n2 = 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] → [n0, n, m, 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%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ n0 < n → (m * n + n0 = m0 * n ↔ m = m0 ∧ n0 = 0); (ws)%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ 0 * n = 0; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n; (elim) Used terms: n0, n, m, n1