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Proof step 19 from proof max_prime_divisor-ge2 (previous step)

max_prime_divisor(i) * i0 = i, 0 < i, i0 < i, 2 ≤ i, 0 ≤ i, i0 ≤ 0 ⊦ i0 = 0, prime(i), 0 ≤ max_prime_divisor(i) ∧ i0 < 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: max_prime_divisor-nls-zero: ⊦ ¬ prime(i) ∧ 2 ≤ i → ¬ max_prime_divisor(i) < 0;% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)