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Proof step 5 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 6:
0 = i + j ⊦ i = j, i = abs(j), i < 0

Rule argument

Additional information

Used simplifier rules: Abs-pos: ⊦ ¬ j < 0 → abs(j) = j;%absneg-01: ⊦ abs(i) ≠ i → abs(i) = ~ i;%Minus-def: ⊦ ~ i = 0 - i;%sub-zero: ⊦ i + (0 - i0) = i - i0;%sub-sub-03: ⊦ i - i = 0;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2