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Proof step 14 from proof gcdex (previous step)

⊦ abs(i) < abs(j), ∃ i0, j0. gcd(abs(i), abs(j)) = i * i0 + j * j0

User applied rule insert lemma

And obtained the following new goals

Proof step 15:
0 ≤ i, 0 ≤ j, j ≤ i ⊦ ∃ i0, j0. gcd(i, j) = i * i0 + j * j0
Proof step 16:
∃ i0, j0. gcd(abs(i), abs(j)) = abs(i) * i0 + abs(j) * j0 ⊦ abs(i) < abs(j), ∃ i0, j0. gcd(abs(i), abs(j)) = i * i0 + j * j0

Rule argument

spec: int-gcd inst: name: gcdhex seq: 0 ≤ i, 0 ≤ j, j ≤ i ⊦ ∃ i0, j0. gcd(i, j) = i * i0 + j * j0 subst: [i, j] → [abs(i), abs(j)] flags: current, []

Additional information

Used simplifier rules: lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%abs-03: ⊦ 0 ≤ abs(i); from specification int-abs Used terms: abs(i), abs(j)