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Proof step 7 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 ⊦ 0 ≤ i0 ∧ j | 1 + ~ j0 * j - 1, ∃ 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: div-neg: ⊦ i | ~ j ↔ i | j;%div-mult: ⊦ i | i * j;%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%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%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%neg-06: ⊦ 0 - i = ~ i; from specification int-abs%num-subadd: ⊦ (* i) - j + (* i0) = i - i0 - j; from specification int-abs%sub-sub: ⊦ i - j - i0 = i - j + i0; from specification int-abs%neg-03: ⊦ ~ i + i = 0; from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%neg-15: ⊦ i + ~ j = i - j; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult