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Proof step 6 from proof max_prime_divisor-ge2 (previous step)

¬ prime(i) ∧ 2 ≤ i → (∃ i0. max_prime_divisor(i) * i0 = i), 2 ≤ i, i / max_prime_divisor(i) < 2, 0 ≤ i, 0 < i, i / max_prime_divisor(i) < i ⊦ prime(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information