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Proof step 3 from proof abs-21 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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