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

m0 % n = 0, n * m / n + n < m0, n * m / n + n < m, m0 ≤ m ⊦ 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] → [m, n, m1, n0]

Additional information

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