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Proof step 2 from proof div2-add-neg (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
i = i0 + 1, j % 2 = 0, abs(i) ≤ j, i ≤ 0 ⊦ (i + j) / 2 = i0 / 2 + j / 2

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, 1, i0] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, 1, i0