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Proof step 14 from proof mod-ls-same-02 (previous step)

i < 0, j + i ≤ 0 ⊦ i < i % j ↔ abs(j) = 1 ∨ ¬ 0 < abs(j) + i, j = 0, j < 0 ∧ i < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
i % j < 0, 0 < j, i < 0, i % j ≤ i, j + i ≤ 0 ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + j < j ↔ i < 0; (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ i ≤ j → (j < i ↔ false); (as)% ⊦ 0 < j → abs(j) = j; (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)