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Proof step 16 from proof prime-smallest (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: m abort: F discard: F computed: F

Additional information

Used simplifier rules: one: ⊦ ¬ prime(1);% ⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n); (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%n ≠ 0 ⊦ (m * n) % n = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → ¬ n < m; (s)% ⊦ m ≠ 0 → (m < 2 ↔ m = 1); (s)% ⊦ m0 * m = 1 ↔ m0 = 1 ∧ m = 1; (ws)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)% ⊦ m < n → n ≠ 0; (s) Used terms: m