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Proof step 20 from proof make (previous step)

i ≠ 0 ⊦ i = 0, rational(i, 1).num = i ∧ rational(i, 1).denom = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ i ≠ 0 ∧ j ≠ 0 → rational(i, j).denom = abs(j) / gcd(i, j)
Proof step 22:
i ≠ 0 ∧ 1 ≠ 0, i ≠ 0 ⊦ i = 0, rational(i, 1).num = i ∧ abs(1) / gcd(i, 1) = 1

Rule argument

name: denominator-pos rotate: F subst: [j, i] → [1, i]

Additional information

Used simplifier rules: numerator-pos: ⊦ i ≠ 0 ∧ 0 < j → rational(i, j).num = i / gcd(i, j); Used terms: 1, i