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Proof step 2 from proof gcd-mod-one (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
0 ≤ j, 0 ≤ i, gcd(i, j) = 1 ⊦ ∃ i0. 0 ≤ i0 ∧ j | i * i0 - 1
Proof step 4:
∃ i0. 0 ≤ i0 ∧ j | i * i0 - 1, gcd(i, j) = 1, 0 < j, 1 < j, 0 < i, 1 < i ⊦ ∃ i0. (i * i0) % j = 1

Rule argument

spec: int-gcd inst: name: gcd-conq-one seq: 0 ≤ j, 0 ≤ i, gcd(i, j) = 1 ⊦ ∃ i0. 0 ≤ i0 ∧ j | i * i0 - 1 subst: [j, i] → [j, i] flags: current, []

Additional information

Used simplifier rules: lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: j, i