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Proof step 1 from proof gcd-divides-both-h (conclusion)
0 ≤ i, i < j, j0 | gcd(i, j) ⊦ j0 | j ∧ j0 | i
System applied rule normalize
And obtained the following new goals
Proof step 2
:
i < j, j0 | gcd(i, j), 0 ≤ i ⊦ j0 | j ∧ j0 | i
Rule argument
Additional information