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Proof step 4 from proof prime-smallest (previous step)

n % m = 0, prime(n), 2 ≤ m, m * m ≤ n ⊦ n = 1, n = 0, m = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

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] → [n, m, m0, n0] flags:

Additional information

Used simplifier rules: mult-03: ⊦ 2 ≤ m ∧ 2 ≤ n → ¬ prime(m * n);%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%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%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%le: ⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n); from specification nat-mult%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n; (elim) Used terms: n, m, m0, n0