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Proof step 5 from proof abs-add-pos (previous step)

i + abs(j) < 0 ⊦ j < 0, i < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i + j < 0 ⊦ j < 0, i < 0

Rule argument

Additional information

Used simplifier rules: Abs-pos: ⊦ ¬ j < 0 → abs(j) = j;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0;