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Proof step 11 from proof mod-ls-same-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
i < 0, j + i ≤ 0 ⊦ i < i % j ↔ abs(j) = 1 ∨ ¬ 0 < abs(j) + i, j = 0

Rule argument

Additional information

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