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Proof step 10 from proof gcd-rec (previous step)

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

User applied rule apply lemma

And obtained the following new goals

Proof step 11:
⊦ i1 | i ∧ i1 | j ∧ 0 ≠ i → i1 ≤ gcd(i, j)
Proof step 12:
abs(i) = i0 * abs(j) + j0, gcd(abs(i), abs(j)) < gcd(abs(j), j0), j0 < abs(j), 0 ≤ i0, 0 ≤ j0, gcd(abs(j), j0) | abs(i) ∧ gcd(abs(j), j0) | abs(j) ∧ 0 ≠ abs(i) → gcd(abs(j), j0) ≤ gcd(abs(i), abs(j)) ⊦ gcd(abs(i), abs(j)) = gcd(abs(j), j0), j = 0

Rule argument

spec: <toplevel> name: gcd-less seq: ⊦ i1 | i ∧ i1 | j ∧ 0 ≠ i → i1 ≤ gcd(i, j) subst: [i1, i, j] → [gcd(abs(j), j0), abs(i), abs(j)]

Additional information

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