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Proof step 10 from proof mod-ls-same-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i < 0, j + i ≤ 0 ⊦ j = 0, i < i % j ↔ abs(j) = 1 ∨ ¬ 0 < abs(j) + i

Rule argument

Additional information

Used simplifier rules: forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs