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Proof step 12 from proof gcd-one (previous step)

⊦ gcd(i, j) = 1, 1 < gcd(i, j) ∧ gcd(i, j) | i ∧ gcd(i, j) | j, ∃ i0. 1 < i0 ∧ i0 | i ∧ i0 | j

User applied rule case distinction

And obtained the following new goals

Proof step 13:
⊦ 1 < gcd(i, j), gcd(i, j) = 1, ∃ i0. 1 < i0 ∧ i0 | i ∧ i0 | j
Proof step 15:
⊦ gcd(i, j) | i, gcd(i, j) = 1, ∃ i0. 1 < i0 ∧ i0 | i ∧ i0 | j
Proof step 18:
⊦ gcd(i, j) | j, gcd(i, j) = 1, ∃ i0. 1 < i0 ∧ i0 | i ∧ i0 | j

Rule argument

pos: right 2

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