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Proof step 4 from proof div2-add-neg (previous step)

j % 2 = 0, i = i0 + 1, i ≤ 0, abs(i) ≤ j ⊦ (i + j) / 2 = i0 / 2 + j / 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
j % 2 = 0, i0 < 0, ~1 ≤ i0 + j ⊦ ((i0 + 1) + j) / 2 = i0 / 2 + j / 2

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; (s)% ⊦ ~ i ≤ j ↔ 0 ≤ i + j; (s)% ⊦ ~ i - j = ~(i + j); (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i < 0 → abs(i + 1) = abs(i) - 1; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i ≤ ~1 ↔ i < 0; (s)% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j; (s)