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

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;% ⊦ ¬ i < 0 ∧ ¬ j < 0 → (0 = i + j ↔ i = 0 ∧ j = 0); (ws)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2