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Proof step 24 from proof congmod-inverse (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 25:
0 ≤ i, (abs(i) * i0) % abs(j) = 1, gcd(i, j) = 1, 0 < i0 ⊦ j | i * i0 - 1, ∃ i0. ≡mod(i * i0, 1, j)
Proof step 31:
(abs(i) * i0) % abs(j) = 1, gcd(i, j) = 1, 0 < i0 ⊦ 0 ≤ i, j | i * i0 - 1, ∃ i0. ≡mod(i * i0, 1, j)

Rule argument

fma: 0 ≤ i

Additional information

Used formulas: 0 ≤ i