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Proof step 5 from proof max_prime_divisor-nls-01 (previous step)

1 < max_prime_divisor(i), 0 < max_prime_divisor(i), i < max_prime_divisor(i), 2 < max_prime_divisor(i), 0 < i, 2 ≤ i, 0 ≤ i, ¬ prime(i) ∧ 2 ≤ i → ¬ i ≤ max_prime_divisor(i) ⊦ prime(i)

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 8

Additional information

Used simplifier rules: ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s)