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

~1 = j2 / gcd(i2, j2) ⊦ j2 + gcd(i2, j2) = 0, 1 = ~ j2 / gcd(i2, j2)

User applied rule apply lemma

And obtained the following new goals

Proof step 7:
⊦ 0 ≤ gcd(i, j)
Proof step 8:
~1 = j2 / gcd(i2, j2), 0 ≤ gcd(i2, j2) ⊦ j2 + gcd(i2, j2) = 0, 1 = ~ j2 / gcd(i2, j2)

Rule argument

spec: <toplevel> name: gcd-pos seq: ⊦ 0 ≤ gcd(i, j) subst: [i, j] → [i2, j2]

Additional information

Used terms: i2, j2