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Proof step 7 from proof abs-shift (previous step)

¬ j < 0, i % j < 0 ⊦ j = 0, i % j + abs(i) % abs(j) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: mod-less-01: ⊦ 0 < j → i % j < j;%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%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%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1