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Proof step 2 from proof gcd-conq-one (previous step)

gcd(i, j) = 1, 0 ≤ i, 0 ≤ j ⊦ ∃ i0. 0 ≤ i0 ∧ j | i * i0 - 1

User applied rule apply lemma

And obtained the following new goals

Proof step 3:
⊦ ∃ i0, j0. gcd(i, j) = i * i0 + j * j0
Proof step 4:
gcd(i, j) = 1, 0 ≤ j, 0 ≤ i, ∃ i0, j0. gcd(i, j) = i * i0 + j * j0 ⊦ ∃ i0. 0 ≤ i0 ∧ j | i * i0 - 1

Rule argument

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

Additional information

Used terms: i, j