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Proof step 4 from proof gcdex (previous step)

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

User applied rule apply lemma

And obtained the following new goals

Proof step 5:
0 ≤ i, 0 ≤ j, j ≤ i ⊦ ∃ i0, j0. gcd(i, j) = i * i0 + j * j0
Proof step 6:
abs(i) < abs(j), ∃ i0, j0. gcd(abs(j), abs(i)) = abs(j) * i0 + abs(i) * j0 ⊦ ∃ i0, j0. gcd(abs(i), abs(j)) = i * i0 + j * j0

Rule argument

spec: <toplevel> name: gcdhex seq: 0 ≤ i, 0 ≤ j, j ≤ i ⊦ ∃ i0, j0. gcd(i, j) = i * i0 + j * j0 subst: [i, j] → [abs(j), abs(i)]

Additional information

Used simplifier rules: ⊦ 0 ≤ abs(i); (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s) Used terms: abs(j), abs(i)