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Proof step 1 from proof num-gcd-pos (conclusion)
⊦ 0 < j ∧ j < i → gcd((* i), (* j)) = gcd(j, i % j)
Special step normalize
And obtained the following new goals
Proof step 2
:
⊦ 0 < j ∧ j < i → gcd(i, j) = gcd(j, i % j)