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Proof step 13 from proof prime-smallest (previous step)

n = m0 * m + n0, n0 = 0, n0 < m, prime(m), m * m ≤ n, ∀ m. 2 ≤ m ∧ m * m ≤ n → n % m ≠ 0 ⊦ m = 0, n = 1, n = 0, prime(n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
prime(m), m ≤ m0, ∀ m1. 2 ≤ m1 ∧ m1 * m1 ≤ m0 * m → (m0 * m) % m1 ≠ 0 ⊦ m0 * m = 1, m = 0, m0 = 0, m0 = 1

Rule argument

Additional information

Used simplifier rules: mult-02: ⊦ prime(m) → (prime(m * n) ↔ n = 1);% ⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n); (s)% ⊦ 0 < n ↔ n ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0; (ws)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ 0 + m = m; (s)