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Proof step 16 from proof abs-pos (previous step)

0 < j ⊦ i = 0, j = 0, abs(rational(i, j)).num = rational(abs(i), j).num ∧ j / gcd(i, j) = j / gcd(abs(i), j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ gcd(abs(i), j) = gcd(i, j)
Proof step 18:
0 < j ⊦ i = 0, j = 0, abs(rational(i, j)).num = rational(abs(i), j).num ∧ j / gcd(i, j) = j / gcd(i, j)

Rule argument

seq: ⊦ gcd(abs(i), j) = gcd(i, j) subst: [j, i] → [j, i] rotate: F spec: int-gcd inst: name: gcd-abs-rew-01

Additional information

Used terms: j, i