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Proof step 5 from proof congmod-inverse (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ gcd(i, j) = gcd(abs(i), abs(j))
Proof step 7:
gcd(abs(i), abs(j)) = 1 ⊦ 1 < abs(j) ∧ 1 < abs(i) ∧ gcd(abs(i), abs(j)) = 1, ∃ i0. ≡mod(i * i0, 1, j)

Rule argument

spec: int-gcd inst: name: gcd-abs-rew-02 seq: ⊦ gcd(i, j) = gcd(abs(i), abs(j)) subst: [j, i] → [j, i] flags: rewrite

Additional information

Used terms: j, i