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Proof step 17 from proof not-prime (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
n0 < m, 2 < m * m, 2 ≤ m * n0, ∀ n1. n1 < m * n0 → ¬ prime(n1) ∧ 2 ≤ n1 → (∃ m. m | n1 ∧ prime(m) ∧ m * m ≤ n1) ⊦ n0 = 0, m = 1, n0 = 1, m * n0 = 1, m = 0, prime(n0), ∃ m0. m0 | m * n0 ∧ prime(m0) ∧ m0 * m0 ≤ m * n0

Rule argument

Additional information

Used simplifier rules: divi-mult: ⊦ m | m * n0;%lels-03: ⊦ m < n → m ≤ n; from specification nat%le: ⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n); from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult