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Proof step 38 from proof congmod-inverse (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
gcd(i, j) = 1, ~1 = (i * i0) % abs(j), 0 < i0, i < 0, i < i0 ⊦ j = 0, j | i * i0 - 1, j | i0 * i + 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%abs-neg: ⊦ abs(~ i) = abs(i); from specification int-abs%subsub: ⊦ ~ i - j = ~(i + j); from specification int-abs