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Proof step 19 from proof congmod-inverse (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
(abs(i) * i0) % abs(j) = 1, gcd(i, j) = 1, 0 < i0 ⊦ ∃ i0. ≡mod(i * i0, 1, j)

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs