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Proof step 11 from proof max_prime_divisor-divides-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: divides-prime seq: ⊦ prime(j) → (i | j ↔ abs(i) = j ∨ abs(i) = 1) subst: [j, i] → [max_prime_divisor(i), j] flags: rewrite, current

Additional information

Used simplifier rules: max_prime_divisor-prime: ⊦ ¬ prime(i) ∧ 2 ≤ i → prime(max_prime_divisor(i));%max_prime_divisor-prime: ⊦ ¬ prime(i) ∧ 2 ≤ i → prime(max_prime_divisor(i));%prime-abs: ⊦ prime(j) → abs(j) = j;%prime-one: ⊦ i ≤ 1 → ¬ prime(i); Used terms: max_prime_divisor(i), j