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Proof step 6 from proof mod-eq-01 (previous step)

n0 < n, m0 < n ⊦ ((m1 * n + n0) + m0) % n = m0 ↔ n0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
n ≠ 0 ⊦ (m + n0 * n) % n = m % n
Proof step 8:
m0 < n, n0 < n ⊦ (m0 + n0) % n = m0 ↔ n0 = 0, n = 0

Rule argument

spec: <toplevel> name: mod-mult seq: n ≠ 0 ⊦ (m + n0 * n) % n = m % n subst: [m, n, n0] → [m0 + n0, n, m1] flags: <all paths>

Additional information

Used simplifier rules: mod-eq: ⊦ m < n → m % n = m;% ⊦ 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% ⊦ n + 0 = n; (s)% ⊦ ¬ m < 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m0 + n0, n, m1