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Proof step 22 from proof gcd-rec (previous step)

abs(i) = i0 * abs(j) + j0, j0 < abs(j), gcd(abs(i), abs(j)) < gcd(abs(j), j0), gcd(abs(j), j0) | abs(j), 0 ≤ j0, 0 ≤ i0 ⊦ gcd(abs(i), abs(j)) = gcd(abs(j), j0), i = 0, j = 0, gcd(abs(j), j0) | abs(i)

User applied rule apply lemma

And obtained the following new goals

Proof step 23:
⊦ gcd(i, j) | j
Proof step 24:
abs(i) = i0 * abs(j) + j0, gcd(abs(i), abs(j)) < gcd(abs(j), j0), j0 < abs(j), gcd(abs(j), j0) | abs(j), gcd(abs(j), j0) | j0, 0 ≤ i0, 0 ≤ j0 ⊦ gcd(abs(i), abs(j)) = gcd(abs(j), j0), i = 0, j = 0, gcd(abs(j), j0) | abs(i)

Rule argument

spec: <toplevel> name: gcd-div2 seq: ⊦ gcd(i, j) | j subst: [i, j] → [abs(j), j0]

Additional information

Used terms: abs(j), j0