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Proof step 43 from proof gcd-rec (previous step)

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

User applied rule apply lemma

And obtained the following new goals

Proof step 44:
⊦ gcd(i, j) | i
Proof step 45:
abs(i) = i0 * abs(j) + j0, gcd(abs(j), j0) < gcd(abs(i), abs(j)), j0 < abs(j), gcd(abs(i), abs(j)) | abs(i), 0 ≤ i0, 0 ≤ j0 ⊦ j = 0, gcd(abs(i), abs(j)) | j0

Rule argument

spec: <toplevel> name: gcd-div1 seq: ⊦ gcd(i, j) | i subst: [i, j] → [abs(i), abs(j)]

Additional information

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