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Proof step 1 from proof abs-21 (conclusion)

⊦ i ≠ j → (abs(i) = abs(j) ↔ i = ~ j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ i = j, abs(i) = abs(j) ↔ 0 = i + j

Rule argument

Additional information

Used simplifier rules: Minus-def: ⊦ ~ i = 0 - i;%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j;