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Proof step 9 from proof div2-add-neg (previous step)

¬ (i0 + j) + 1 ≠ 0, j % 2 = 0, ~1 ≤ i0 + j, i0 < 0, j < 0 ⊦ ((i0 + 1) + j) / 2 = i0 / 2 + j / 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
~1 = i0 + j, j % 2 = 0, j < 0, i0 < 0 ⊦ ((i0 + 1) + j) / 2 = i0 / 2 + j / 2

Rule argument

Additional information

Used simplifier rules: forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs