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Proof step 2 from proof gcd-conq-one-01 (previous step)

gcd(i, j) = 1 ⊦ ∃ i0. j | i * i0 - 1

User applied rule insert lemma

And obtained the following new goals

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

Rule argument

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

Additional information

Used terms: i, j