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Proof step 4 from proof prime-def-mod (previous step)

0 < j, 1 < j ⊦ 2 ≤ j ∧ (∀ i. 1 < i ∧ i < j → ¬ i | j) ↔ (∀ i. 1 < i ∧ i * i ≤ j → j % i ≠ 0)

User applied rule case distinction

And obtained the following new goals

Proof step 5:
2 ≤ j ∧ (∀ i. 1 < i ∧ i < j → ¬ i | j), 0 < j, 1 < j ⊦ ∀ i. 1 < i ∧ i * i ≤ j → j % i ≠ 0
Proof step 12:
∀ i. 1 < i ∧ i * i ≤ j → j % i ≠ 0, 0 < j, 1 < j ⊦ 2 ≤ j ∧ (∀ i. 1 < i ∧ i < j → ¬ i | j)

Rule argument

pos: right 1

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