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Proof step 17 from proof congmod-inverse (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 18:
gcd(abs(i), abs(j)) = 1, abs(j) ≤ 1 ⊦ j = 0, ∃ i0. ≡mod(i * i0, 1, j)

Rule argument

fma: j = 0

Additional information

Used simplifier rules: cong-zero: ⊦ ≡mod(i, j, 0) ↔ i = j;%gcd-zero-04: ⊦ gcd(i, 0) = abs(i); from specification int-gcd%abs-01: ⊦ abs(abs(i)) = abs(i); from specification int-abs%eq-one: ⊦ i * j = 1 ↔ abs(i) = 1 ∧ i = j; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used formulas: j = 0