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Proof step 15 from proof is-aligned-still-space (previous step)

n * n1 < (n1 * m) % (n * n1) + n1, m < n ⊦ (n1 * m) % (n * n1) = 0, n1 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ m < n → m % n = m

Rule argument

spec: nat-div name: mod-eq seq: ⊦ m < n → m % n = m subst: [n, m] → [n * n1, n1 * m] flags: <all paths>

Additional information

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