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Proof step 3 from proof absa (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: abs-08: ⊦ i < 0 → abs(i) = ~ i;%Minus-def: ⊦ ~ i = 0 - i;%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0;