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Proof step 18 from proof gcd-rec (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 19:
⊦ gcd(i, j) | i
Proof step 20:
gcd(abs(j), j0) | abs(j), abs(i) = i0 * abs(j) + j0, j0 < abs(j), 0 ≤ j0, 0 ≤ i0, gcd(i0 * abs(j) + j0, abs(j)) < gcd(abs(j), j0) ⊦ gcd(i0 * abs(j) + j0, abs(j)) = gcd(abs(j), j0), j = 0, i = 0, gcd(abs(j), j0) | i0 * abs(j) + j0

Rule argument

spec: int-gcd inst: name: gcd-div1 seq: ⊦ gcd(i, j) | i subst: [i, j] → [abs(j), j0] flags: current, []

Additional information

Used terms: abs(j), j0