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Proof step 23 from proof congmod-inverse (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
(abs(i) * i0) % abs(j) = 1, gcd(i, j) = 1, 0 < i0 ⊦ j | i * i0 - 1, ∃ i0. ≡mod(i * i0, 1, j)

Rule argument

Additional information

Used simplifier rules: div-abs-02: ⊦ i | abs(j) ↔ i | j; from specification int-gcd