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Proof step 3 from proof le-succ (previous step)

m < m / (n + 1) * n ⊦ m / (n + 1) = 0, n = 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n

Rule argument

spec: <toplevel> name: div-elim seq: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n subst: [m, n, m0, n0] → [m, n + 1, m0, n0]

Additional information

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