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Proof step 19 from proof not-prime (previous step)

2 < m * m, n0 < m, 2 ≤ m * n0, n0 ≤ m, ∀ 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 apply induction

And obtained the following new goals

Proof step 20:
n0 < m, 2 < m * m, n0 ≤ m, 2 ≤ m * n0, prime(n0) ∨ (∃ m. m | n0 ∧ prime(m) ∧ m * m ≤ n0), ∀ n1. n1 < m * n0 → 2 ≤ n1 → prime(n1) ∨ (∃ m. m | n1 ∧ prime(m) ∧ m * m ≤ n1) ⊦ m * n0 = 1, m = 1, n0 = 1, n0 = 0, prime(n0), ∃ m0. m0 | m * n0 ∧ prime(m0) ∧ m0 * m0 ≤ m * n0

Rule argument

subst: [n1] → [n0]

Additional information

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