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Proof step 33 from proof congmod-inverse (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
gcd(i, j) = 1, ~1 = (i * i0) % abs(j), 0 < i0, i < 0, i < i0 ⊦ j = 0, j | i * i0 - 1, ∃ i0. ≡mod(i * i0, 1, j)

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j); from specification int-div%abs-07: ⊦ abs(i) = 0 ↔ i = 0; from specification int-abs