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Proof step 2 from proof congmod-inverse (previous step)

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

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 3:
1 < j, 1 < i, gcd(i, j) = 1 ⊦ ∃ i0. (i * i0) % j = 1 ∧ 0 < i0
Proof step 4:
gcd(i, j) = 1 ⊦ 1 < abs(j) ∧ 1 < abs(i) ∧ gcd(abs(i), abs(j)) = 1, ∃ i0. ≡mod(i * i0, 1, j)
Proof step 19:
∃ i0. (abs(i) * i0) % abs(j) = 1 ∧ 0 < i0, gcd(i, j) = 1 ⊦ ∃ i0. ≡mod(i * i0, 1, j)

Rule argument

spec: int-gcd inst: name: gcd-mod-one-01 seq: 1 < j, 1 < i, gcd(i, j) = 1 ⊦ ∃ i0. (i * i0) % j = 1 ∧ 0 < i0 subst: [j, i] → [abs(j), abs(i)] flags:

Additional information

Used terms: abs(j), abs(i)