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Proof step 15 from proof prime-smallest (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
m < m0, prime(m), 2 ≤ m0, ∀ m1. 2 ≤ m1 ∧ m1 * m1 ≤ m0 * m → (m0 * m) % m1 ≠ 0 ⊦ m0 * m = 1, m = m0, m = 0, m0 = 1

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m ≠ 1 → (m < 2 ↔ m = 0); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%n ≠ 0 ⊦ (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% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n); (s)