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Proof step 7 from proof mod-eq-01 (previous step)

n ≠ 0, (m0 + n0) % n = m0, m0 < n, n0 < n ⊦ n = 0, n0 = 0

User applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: nat-div inst: name: div-elim seq: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n subst: [m, n, m0, n0] → [m0 + n0, n, m, n1] flags: rewrite, current

Additional information

Used simplifier rules: eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); 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%ls-01: ⊦ m0 * m < m ↔ m0 = 0 ∧ m ≠ 0; from specification nat-mult Used terms: m0 + n0, n, m, n1