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Proof step 1 from proof rax (conclusion)

⊦ prime(n) ↔ ¬ ¬(1 < n ∧ (∀ m. m | n → m = 1 ∨ m = n))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
1 < n, m | n, ∀ m. m | n → m = 1 ∨ m = n ⊦ m = 1, m = n

Rule argument

Additional information

Used simplifier rules: Prime: ⊦ prime(n) ↔ 1 < n ∧ (∀ m. m | n → m = 1 ∨ m = n);