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Proof step 8 from proof gcd-conq-one (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 9:
i = 0, 1 = i * i0 + j0 * j, gcd(i, j) = 1, i0 < 0, 0 ≤ i, 0 ≤ j ⊦ ∃ i0. 0 ≤ i0 ∧ j | i * i0 - 1
Proof step 11:
1 = i * i0 + j0 * j, gcd(i, j) = 1, i0 < 0, 0 ≤ i, 0 ≤ j ⊦ i = 0, ∃ i0. 0 ≤ i0 ∧ j | i * i0 - 1

Rule argument

fma: i = 0

Additional information

Used formulas: i = 0