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Proof step 11 from proof congmod-inverse (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 12:
i = 1, gcd(abs(i), abs(j)) = 1, abs(i) < abs(j), abs(i) ≤ 1, 1 < abs(j) ⊦ i = 0, j = 0, ∃ i0. ≡mod(i * i0, 1, j)
Proof step 14:
gcd(abs(i), abs(j)) = 1, abs(i) < abs(j), abs(i) ≤ 1, 1 < abs(j) ⊦ i = 1, i = 0, j = 0, ∃ i0. ≡mod(i * i0, 1, j)

Rule argument

fma: i = 1

Additional information

Used formulas: i = 1