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Proof step 4 from proof gcd-mod-one-01 (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 5:
(i * i0) % j = 1, gcd(i, j) = 1, 0 < j, 1 < j, 0 < i, 1 < i ⊦ (i * i0) % j = 1 ∧ 0 < i0

Rule argument

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

Additional information

Used terms: i0