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Proof step 5 from proof gcd-mod-one (previous step)

gcd(i, j) = 1, j | i * i0 - 1, 0 < j, 1 < j, 0 < i, 1 < i, 0 ≤ i0 ⊦ ∃ i0. (i * i0) % j = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 7:
gcd(i, j) = 1, ∃ i1. j * i1 = i * i0 - 1, 0 < j, 1 < j, 0 < i, 1 < i, 0 ≤ i0 ⊦ ∃ i0. (i * i0) % j = 1

Rule argument

spec: int-gcd inst: name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [i * i0 - 1, j] flags: rewrite, current

Additional information

Used terms: i * i0 - 1, j