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Proof step 15 from proof not-prime (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
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, ∃ m0. m0 | m * n0 ∧ prime(m0) ∧ m0 * m0 ≤ m * n0

Rule argument

Additional information

Used simplifier rules: not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%ls-02: ⊦ m0 ≠ 0 → (m0 * m < m0 * n ↔ m < n); from specification nat-mult%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); 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%eq-same: ⊦ m * n = m ↔ ¬ ¬(m = 0 ∨ n = 1); from specification nat-mult