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Proof step 8 from proof div2-add-neg (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ i + 1 ≠ 0 → (~1 ≤ i ↔ 0 ≤ i)
Proof step 10:
j % 2 = 0, (i0 + j) + 1 = 0, j < 0, i0 < 0, ~1 ≤ i0 + j ⊦ ((i0 + 1) + j) / 2 = i0 / 2 + j / 2

Rule argument

spec: int-abs name: mone-07 seq: ⊦ i + 1 ≠ 0 → (~1 ≤ i ↔ 0 ≤ i) subst: [i] → [i0 + j] flags: <all paths>

Additional information

Used simplifier rules: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i < 0 ∧ j < 0 → ¬ 0 ≤ i + j; (s) Used terms: i0 + j