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Proof step 13 from proof modmod-add-if3 (previous step)

¬ 0 ≤ i, 0 < j ⊦ (i0 < 0 ⊃ (i + i0) % j;(i + i0 < 0 ⊃ (i + i0) % j;(0 ≤ i + i0 % j ⊃ (i + i0) % j;((i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j - j;0)))) = (0 ≤ i + i0 % j ∧ i + i0 < 0 ∧ 0 ≤ i ∧ i0 < 0 ∧ (i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j + j ;(i + i0 % j < 0 ∧ 0 ≤ i + i0 ∧ i < 0 ∧ 0 ≤ i0 ∧ (i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j - j;(i + i0) % j) )

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs