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Proof step 10 from proof gcd-one (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 11:
⊦ gcd(i, j) = 1, ∃ i0. 1 < i0 ∧ i0 | i ∧ i0 | j, 1 < gcd(i, j) ∧ gcd(i, j) | i ∧ gcd(i, j) | j

Rule argument

pos: right 2 terms: gcd(i, j) abort: F discard: F computed: F

Additional information

Used terms: gcd(i, j)