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Proof step 6 from proof absa (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i < 0, i + j < 0 ⊦ 0 = abs(j) + j, 0 < j

Rule argument

Additional information

Used simplifier rules: Abs-neg: ⊦ j < 0 → abs(j) = 0 - j;%sub-add-01: ⊦ i + (j - i0) = i + j - i0;%sub-add: ⊦ i + j - i = j;% ⊦ i < j → ¬ j < i; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ 0 + i = i; (s)