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Proof step 10 from proof gcd-conq-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i0 < 0 ⊦ ∃ i. 0 ≤ i

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;% ⊦ i * j = 1 ↔ abs(i) = 1 ∧ i = j; (ws)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + 0 = i; (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ i * 0 = 0; (s)% ⊦ 0 ≤ i → abs(i) = i; (s)