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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 ⊦ m = 0, m0 * m = 1, m0 = 0, m0 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat%twols-02: ⊦ m ≠ 0 → (m < 2 ↔ m = 1); from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%mod-mult-01: n ≠ 0 ⊦ (m * n) % n = 0; from specification nat-div%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%le: ⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n); from specification nat-mult