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Proof step 3 from proof gcd-conq-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
1 = i * i0 + j * j0, gcd(i, j) = 1, 0 ≤ j, 0 ≤ i ⊦ ∃ i0. 0 ≤ i0 ∧ j | i * i0 - 1

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs