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Proof step 15 from proof gcd-conq-one (previous step)

gcd(i, j) = 1, 1 = i * i0 + j0 * j, i0 < 0, 0 < i, i0 + 1 < i, 0 ≤ j ⊦ i = 0, ∃ i0. 0 ≤ i0 ∧ j | i * i0 - 1

User applied rule exists right

And obtained the following new goals

Proof step 16:
gcd(i, j) = 1, 1 = i * i0 + j0 * j, i0 < 0, 0 < i, i0 + 1 < i, 0 ≤ j ⊦ i = 0, 0 ≤ j * j0 + i0 ∧ j | i * (j * j0 + i0) - 1

Rule argument

pos: right 2 terms: j * j0 + i0 abort: F discard: T computed: F

Additional information

Used terms: j * j0 + i0