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Proof step 6 from proof gcd-one-eq (previous step)

j2 / gcd(i2, j2) = 1, 0 ≤ gcd(i2, j2), 0 ≤ j2 ⊦ gcd(i2, j2) = j2

User applied rule insert lemma

And obtained the following new goals

Proof step 7:
⊦ i | j ∧ i ≠ 0 → (j / i = 1 ↔ j = i)
Proof step 8:
gcd(i2, j2) | j2 ∧ gcd(i2, j2) ≠ 0 → (j2 / gcd(i2, j2) = 1 ↔ j2 = gcd(i2, j2)), j2 / gcd(i2, j2) = 1, 0 ≤ gcd(i2, j2), 0 ≤ j2 ⊦ gcd(i2, j2) = j2

Rule argument

spec: int-gcd inst: name: div-eq-one seq: ⊦ i | j ∧ i ≠ 0 → (j / i = 1 ↔ j = i) subst: [i, j] → [gcd(i2, j2), j2] flags: current, []

Additional information

Used terms: gcd(i2, j2), j2