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Proof step 9 from proof gcd-conq-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%div-one: ⊦ 1 | i;%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%eq-one: ⊦ i * j = 1 ↔ abs(i) = 1 ∧ i = j; from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-left: ⊦ 0 * i = 0; from specification int-mult