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

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

User applied rule insert lemma

And obtained the following new goals

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

Rule argument

spec: int-mod inst: name: congmod-inverse seq: ⊦ gcd(i, j) = 1 → (∃ i0. ≡mod(i * i0, 1, j)) subst: [i, j] → [abs(i), abs(j)] flags: current, []

Additional information

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