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Proof step 3 from proof abs-add-zero (previous step)

abs(i) + abs(j) = 0 ⊦ i < 0, i = 0 ∧ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
i + abs(j) = 0 ⊦ i < 0, i = 0 ∧ j = 0

Rule argument

Additional information

Used simplifier rules: Abs-pos: ⊦ ¬ j < 0 → abs(j) = j;