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Proof step 4 from proof mod-ls-same-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i < 0 ⊦ i < i % j ↔ abs(j) = 1 ∨ abs(j) ≤ ~ i, j = 0

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i < 0 → abs(i) = ~ i; (s)