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Proof step 13 from proof gcd-conq-one (previous step)

1 = i * i0 + j0 * j, gcd(i, j) = 1, i0 < 0, 0 < i, 0 ≤ j ⊦ i = 0, ∃ i0. 0 ≤ i0 ∧ j | i * i0 - 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ i ≤ j ↔ ¬ j < i
Proof step 15:
1 = i * i0 + j0 * j, gcd(i, j) = 1, 0 < i, i0 < 0 ⊦ i = 0, j < 0, ∃ i0. 0 ≤ i0 ∧ j | i * i0 - 1

Rule argument

spec: int-abs name: le-def seq: ⊦ i ≤ j ↔ ¬ j < i subst: [j, i] → [j, 0] flags: <all paths>

Additional information

Used terms: j, 0