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Proof step 9 from proof congmod-inverse (previous step)

¬ abs(j) ≤ 1, gcd(abs(i), abs(j)) = 1 ⊦ ∃ i0. ≡mod(i * i0, 1, j), 1 < abs(j) ∧ 1 < abs(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0; from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lel: ⊦ i < i0 ∧ i = i1 → i1 < i0; (forward) from specification oelem[intlista]%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%numlelsdel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* i0) < j; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs