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Proof step 21 from proof gcd-rec (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 22:
⊦ gcd(i, j) | j
Proof step 23:
gcd(abs(j), j0) | j0, abs(i) = i0 * abs(j) + j0, gcd(abs(j), j0) | abs(j), j0 < abs(j), 0 ≤ j0, 0 ≤ i0, gcd(abs(i), abs(j)) < gcd(abs(j), j0) ⊦ gcd(abs(i), abs(j)) = gcd(abs(j), j0), j = 0, i = 0, gcd(abs(j), j0) | abs(i)

Rule argument

spec: int-gcd inst: name: gcd-div2 seq: ⊦ gcd(i, j) | j subst: [i, j] → [abs(j), j0] flags: current, []

Additional information

Used terms: abs(j), j0