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Proof step 13 from proof abs-pos (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ i ≠ 0 ∧ j ≠ 0 → rational(i, j).denom = abs(j) / gcd(i, j)
Proof step 15:
abs(i) ≠ 0 ∧ j ≠ 0, 0 < j ⊦ j = 0, i = 0, abs(rational(i, j)).num = rational(abs(i), j).num ∧ j / gcd(i, j) = abs(j) / gcd(abs(i), j)

Rule argument

name: denominator-pos rotate: F subst: [j, i] → [j, abs(i)]

Additional information

Used simplifier rules: ⊦ abs(i) = 0 ↔ i = 0; Used terms: j, abs(i)