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Proof step 2 from proof zero-06 (previous step)

⊦ 0 = i + j ↔ i = 0 ∧ j = 0, j < 0, i < 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;%zero-05: ⊦ 0 < i ∧ ¬ j < 0 → 0 ≠ i + j;% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)