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Proof step 25 from proof congmod-inverse (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
gcd(i, j) = 1, (i * i0) % abs(j) = 1, 0 < i0, 0 ≤ i ⊦ j | i * i0 - 1, ∃ i0. ≡mod(i * i0, 1, j)

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs