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

m % (n1 * n0) = 0, n + 1 = n1 * n0 ⊦ m % n0 = 0, n0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ n ≠ 0 → (m % n = 0 ↔ (∃ n0. m = n0 * n))

Rule argument

spec: <toplevel> name: modex seq: ⊦ n ≠ 0 → (m % n = 0 ↔ (∃ n0. m = n0 * n)) subst: [n, m] → [n1 * n0, m] flags: <all paths>

Additional information

Used simplifier rules: mod-mult-01: n ≠ 0 ⊦ (m * n) % n = 0;% ⊦ 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% ⊦ 0 * n = 0; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n1 * n0, m