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Proof step 2 from proof mod-once (previous step)

⊦ (m + n) % n = m % n, n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
n ≠ 0, m / n * n ≤ m ⊦ m % n = m - m / n * n
Proof step 4:
n ≠ 0 ∧ (m + n) / n * n ≤ m + n ⊦ m + n - (m + n) / n * n = m % n, n = 0

Rule argument

spec: nat-div inst: name: rax seq: n ≠ 0, m / n * n ≤ m ⊦ m % n = m - m / n * n subst: [m, n] → [m + n, n] flags: rewrite, current

Additional information

Used simplifier rules: add: n ≠ 0 ⊦ (n0 + n) / n = n0 / n + 1;%cond-01: n ≠ 0 ⊦ ¬ m < m / n * n;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%rsucc-02: ⊦ m * (n + 1) = m * n + m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat Used terms: m + n, n