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Proof step 11 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 = max_prime_divisor(i), j ∈ all_prime_divisors(i / max_prime_divisor(i)) ↔ prime(j) ∧ j | i, prime(i)

User applied rule apply induction

And closed the goal!

Rule argument

subst: [i0, j] → [i / max_prime_divisor(i), j]

Additional information

Used simplifier rules: max_prime_divisor-ge2: ⊦ ¬ prime(i) ∧ 2 ≤ i → 2 ≤ i / max_prime_divisor(i); from specification int-gcd%max_prime_divisor-divides-02: ⊦ ¬ j | i ∧ prime(j) ∧ ¬ prime(i) ∧ 2 ≤ i → ¬ j | i / max_prime_divisor(i); from specification int-gcd%max_prime_divisor-divides-03: ⊦ ¬ prime(i) ∧ 2 ≤ i ∧ prime(j) ∧ j | i ∧ j ≠ max_prime_divisor(i) → j | i / max_prime_divisor(i); from specification int-gcd%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%max_prime_divisor-abs: ⊦ ¬ prime(i) ∧ 2 ≤ i → abs(i / max_prime_divisor(i)) = i / max_prime_divisor(i); from specification int-gcd%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%max_prime_divisor-nle-01: ⊦ ¬ prime(i) ∧ 2 ≤ i → ¬ i ≤ i / max_prime_divisor(i); from specification int-gcd%max_prime_divisor-not-zero: ⊦ ¬ prime(i) ∧ 2 ≤ i → max_prime_divisor(i) ≠ 0; from specification int-gcd%abs: j ≠ 0 ⊦ abs(i / j) = abs(i) / abs(j); from specification int-div%max_prime_divisor-abs-01: ⊦ ¬ prime(i) ∧ 2 ≤ i → abs(max_prime_divisor(i)) = max_prime_divisor(i); from specification int-gcd Used terms: i / max_prime_divisor(i), j