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Proof step 2 from proof abs-12 (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: absle: ⊦ 0 ≤ i → abs(i) = i;%abs-18: ⊦ abs(i) = i ↔ 0 ≤ i;%le-def: ⊦ i ≤ j ↔ ¬ j < i;%zero-gs: ⊦ 0 ≤ i ∧ 0 ≤ j → ¬ i + j < 0;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2