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Proof step 20 from proof congmod-inverse (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 21:
(abs(i) * i0) % abs(j) = 1, gcd(i, j) = 1, 0 < i0 ⊦ ≡mod(i * i0, 1, j), ∃ i0. ≡mod(i * i0, 1, j)

Rule argument

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

Additional information

Used terms: i0