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Proof step 11 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
gcd(i, j) = 1, 1 = i * i0 + j0 * j, i0 < 0, 0 ≤ j, 0 < i, i0 + 1 < i ⊦ i = 0, ∃ i0. 0 ≤ i0 ∧ j | i * i0 - 1

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; from specification int-basic2%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs