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

gcd(i, j) = 1, i * i0 = j * i1 + 1, 0 < j, 1 < j, 0 < i, 1 < i, 0 ≤ i0 ⊦ ∃ i0. (i * i0) % j = 1

User applied rule exists right

And obtained the following new goals

Proof step 9:
gcd(i, j) = 1, i * i0 = j * i1 + 1, 0 < j, 1 < j, 0 < i, 1 < i, 0 ≤ i0 ⊦ (i * i0) % j = 1

Rule argument

pos: right 1 terms: i0 abort: F discard: T computed: F

Additional information

Used terms: i0