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Proof step 6 from proof all_prime_divisors-in-01 (previous step)

prime(i), 2 ≤ i, prime(j), j | 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 = i

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: prime-abs: ⊦ prime(j) → abs(j) = j; from specification int-gcd%prime-one: ⊦ i ≤ 1 → ¬ prime(i); from specification int-gcd Used terms: i, j