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Proof step 1 from proof div2-add-neg (conclusion)

⊦ j % 2 = 0 ∧ i ≤ 0 ∧ abs(i) ≤ j → (i + j) / 2 = (i - 1) / 2 + j / 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
j % 2 = 0, abs(i) ≤ j, i ≤ 0 ⊦ (i + j) / 2 = (i - 1) / 2 + j / 2

Rule argument

Additional information

Used simplifier rules: forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs