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Proof step 44 from proof gcd-rec (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 45:
⊦ gcd(i, j) | j
Proof step 46:
gcd(abs(i), abs(j)) | abs(j), gcd(abs(i), abs(j)) | abs(i), abs(i) = i0 * abs(j) + j0, gcd(abs(j), j0) < gcd(abs(i), abs(j)), 0 ≤ i0, 0 ≤ j0, j0 < abs(j) ⊦ j = 0, gcd(abs(i), abs(j)) | j0

Rule argument

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

Additional information

Used terms: abs(i), abs(j)