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Proof step 36 from proof gcd-rec (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 37:
⊦ i1 | i ∧ i1 | j ∧ 0 ≠ i → i1 ≤ gcd(i, j)
Proof step 38:
gcd(abs(i), abs(j)) | abs(j) ∧ gcd(abs(i), abs(j)) | j0 ∧ 0 ≠ abs(j) → gcd(abs(i), abs(j)) ≤ gcd(abs(j), j0), abs(i) = i0 * abs(j) + j0, gcd(abs(j), j0) < gcd(abs(i), abs(j)), 0 ≤ i0, 0 ≤ j0, j0 < abs(j) ⊦ j = 0

Rule argument

spec: int-gcd inst: name: gcd-less seq: ⊦ i1 | i ∧ i1 | j ∧ 0 ≠ i → i1 ≤ gcd(i, j) subst: [i1, i, j] → [gcd(abs(i), abs(j)), abs(j), j0] flags: current, []

Additional information

Used terms: gcd(abs(i), abs(j)), abs(j), j0