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Proof step 2 from proof divides-prime (previous step)

prime(j) ⊦ i | j ↔ abs(i) = j ∨ abs(i) = 1

User applied rule case distinction

And obtained the following new goals

Proof step 3:
i | j, prime(j) ⊦ abs(i) = j ∨ abs(i) = 1
Proof step 23:
abs(i) = j ∨ abs(i) = 1, prime(j) ⊦ i | j

Rule argument

pos: right 1

Additional information