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Proof step 5 from proof not-prime (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
2 ≤ n, m | n, ∀ n0. n0 < n → ¬ prime(n0) ∧ 2 ≤ n0 → (∃ m. m | n0 ∧ prime(m) ∧ m * m ≤ n0) ⊦ n = 0, m = 1, m = n, n = 1, ∃ m. m | n ∧ prime(m) ∧ m * m ≤ n

Rule argument

Additional information

Used simplifier rules: one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1