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Proof step 11 from proof gcd-one-eq (previous step)

j2 / gcd(i2, j2) = 1, 0 < j2, 0 < gcd(i2, j2) ⊦ gcd(i2, j2) = j2, gcd(i2, j2) = 0, j2 = 0, gcd(i2, j2) | j2

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: gcd-div2 seq: ⊦ gcd(i, j) | j subst: [i, j] → [i2, j2]

Additional information

Used terms: i2, j2