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Proof step 21 from proof not-prime (previous step)

2 < m * m, n0 < m, m0 * m0 < m, prime(m0), m0 | n0, 2 ≤ m * n0, n0 ≤ m, m0 * m0 ≤ n0, ∀ n1. n1 < m * n0 → 2 ≤ n1 → prime(n1) ∨ (∃ m. m | n1 ∧ prime(m) ∧ m * m ≤ n1) ⊦ m * n0 = 1, m = 1, n0 = 0, n0 = 1, prime(n0), ∃ m0. m0 | m * n0 ∧ prime(m0) ∧ m0 * m0 ≤ m * n0

User applied rule exists right

And closed the goal!

Rule argument

pos: right 6 terms: m0 abort: F discard: F computed: F

Additional information

Used simplifier rules: divi-mult-01: ⊦ m | n → m | n * n0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m * m = 0 ↔ m = 0; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ m0 * m < m ↔ m0 = 0 ∧ m ≠ 0; (ws) Used terms: m0