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Proof step 20 from proof ex-modeq-half (previous step)

(i0 % j) % j < 0, i0 % j < 0, j / 2 < abs(i0 % j + j), 0 < j, odd(j) ⊦ ∃ i. ≡mod(i, i0, j) ∧ abs(i) ≤ j / 2, (i0 % j) % j + j = i0 % j + j ∧ 0 ≤ i0 % j + j / 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
i0 % j < 0, j / 2 < abs(i0 % j + j), 0 < j, odd(j), i0 % j + j / 2 < 0, i0 % j + j / 2 < j ⊦ ∃ i. ≡mod(i, i0, j) ∧ abs(i) ≤ j / 2

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]%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%modmod: j ≠ 0 ⊦ (i % j) % j = i % j; from specification int-div%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%neg: ⊦ 0 < ~ i ↔ i < 0; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; 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%ax-04: ⊦ i + 0 = i; from specification int-basic2