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Proof step 31 from proof congmod-inverse (previous step)

(abs(i) * i0) % abs(j) = 1, gcd(i, j) = 1, 0 < i0 ⊦ 0 ≤ i, j | i * i0 - 1, ∃ i0. ≡mod(i * i0, 1, j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
gcd(i, j) = 1, ~(i * i0) % abs(j) = 1, 0 < i0, i < 0, i < i0 ⊦ j | i * i0 - 1, ∃ i0. ≡mod(i * i0, 1, j)

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%transitivity: ⊦ i < i0 ∧ i0 < i1 → i < i1; (forward) from specification oelem[intlista]%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%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs