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Proof step 28 from proof congmod-inverse (previous step)

i * i0 = i1 * abs(j) + 1, gcd(i, j) = 1, 0 < i0, 0 ≤ i, 1 < abs(j), 0 ≤ i1 ⊦ j = 0, j | i * i0 - 1, ∃ i0. ≡mod(i * i0, 1, j)

User applied rule insert equation

And obtained the following new goals

Proof step 29:
i * i0 = i1 * abs(j) + 1, gcd(i, j) = 1, 0 < i0, 0 ≤ i, 1 < abs(j), 0 ≤ i1 ⊦ j = 0, j | i1 * abs(j) + 1 - 1, ∃ i0. ≡mod(i * i0, 1, j)

Rule argument

pos: left 1 flags:

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