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Proof step 4 from proof divides-prime (previous step)

i | j, prime(j) ⊦ abs(i) = j, abs(i) = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ prime(j) ↔ 2 ≤ j ∧ (∀ i. 1 < i ∧ i < j → ¬ i | j)
Proof step 6:
i | j, 2 ≤ j ∧ (∀ i. 1 < i ∧ i < j → ¬ i | j) ⊦ abs(i) = j, abs(i) = 1

Rule argument

spec: int-gcd inst: name: prime-def seq: ⊦ prime(j) ↔ 2 ≤ j ∧ (∀ i. 1 < i ∧ i < j → ¬ i | j) subst: [j] → [j] flags: rewrite, current

Additional information

Used terms: j