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Proof step 8 from proof gcd-conq-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
1 = i * i0 + j0 * j, gcd(i, j) = 1, i0 < 0, 0 ≤ j, 0 ≤ i ⊦ ∃ i0. 0 ≤ i0 ∧ j | i * i0 - 1

Rule argument

Additional information

Used simplifier rules: div-neg: ⊦ i | ~ j ↔ i | j;%div-mult: ⊦ i | i * j;% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ 0 - i = ~ i; (s)% ⊦ (* i) - j + (* i0) = i - i0 - j; (s)% ⊦ i - j - i0 = i - j + i0; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i + ~ j = i - j; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ ~ i + i = 0; (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ ~ i * j = ~(i * j); (s)