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Proof step 5 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 6:
1 < j, 0 < j, 2 ≤ j, ∀ i. 1 < i ∧ i < j → ¬ i | j ⊦ ∀ i. 1 < i ∧ i * i ≤ j → j % i ≠ 0
Proof step 13:
1 < j, 0 < j, ∀ i. 1 < i ∧ i * i ≤ j → j % i ≠ 0 ⊦ 2 ≤ j ∧ (∀ i. 1 < i ∧ i < j → ¬ i | j)

Rule argument

pos: right 1

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