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Proof step 18 from proof congmod-inverse (previous step)

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

User applied rule cut formula

And closed the goal!

Rule argument

fma: j = 1

Additional information

Used simplifier rules: cong-one: ⊦ ≡mod(i, j, 1);%cong-one-01: ⊦ ≡mod(i, j, ~1);%abs-one-01: ⊦ i ≠ 0 ∧ i ≠ 1 → (abs(i) ≤ 1 ↔ i = ~1); from specification int-abs%gcd-one-01: ⊦ gcd(i, 1) = 1; from specification int-gcd Used formulas: j = 1