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Proof step 9 from proof all_prime_divisors-in-01 (previous step)

2 ≤ i, 0 < i, 0 ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → true → (j ∈ all_prime_divisors(i0) ↔ prime(j) ∧ j | i0 ∧ 2 ≤ i0) ⊦ j ∈ all_prime_divisors(i / max_prime_divisor(i)) ∨ j = max_prime_divisor(i) ↔ prime(j) ∧ j | i, prime(i)

User applied rule cut formula

And obtained the following new goals

Proof step 10:
2 ≤ i, 0 < i, 0 ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → true → (j ∈ all_prime_divisors(i0) ↔ prime(j) ∧ j | i0 ∧ 2 ≤ i0) ⊦ j = max_prime_divisor(i), j ∈ all_prime_divisors(i / max_prime_divisor(i)) ∨ j = max_prime_divisor(i) ↔ prime(j) ∧ j | i, prime(i)

Rule argument

fma: j = max_prime_divisor(i)

Additional information

Used simplifier rules: max_prime_divisor-prime: ⊦ ¬ prime(i) ∧ 2 ≤ i → prime(max_prime_divisor(i)); from specification int-gcd%max_prime_divisor-divides: ⊦ ¬ prime(i) ∧ 2 ≤ i → max_prime_divisor(i) | i; from specification int-gcd Used formulas: j = max_prime_divisor(i)