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Proof step 8 from proof congmod-inverse (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 9:
¬ abs(j) ≤ 1, gcd(abs(i), abs(j)) = 1 ⊦ ∃ i0. ≡mod(i * i0, 1, j), 1 < abs(j) ∧ 1 < abs(i)
Proof step 16:
gcd(abs(i), abs(j)) = 1 ⊦ ¬ abs(j) ≤ 1, ∃ i0. ≡mod(i * i0, 1, j), 1 < abs(j) ∧ 1 < abs(i)

Rule argument

fma: ¬ abs(j) ≤ 1

Additional information

Used formulas: ¬ abs(j) ≤ 1