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Proof step 12 from proof div2-nls-02 (previous step)

i + j = i0 * 2 + j0, i0 < j, i + j < 0, i0 < i, abs(j0) < 2, j0 ≤ 0, i0 ≤ 0, i ≤ j ⊦ j0 = ~1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ j < 0 → abs(j) = 0 - j
Proof step 14:
i + j = i0 * 2 + j0, abs(j0) < 2, i0 < i, i + j < 0, i0 < j, i ≤ j, i0 ≤ 0, j0 ≤ 0 ⊦ j0 = ~1, j0 < 0

Rule argument

spec: int-abs name: Abs-neg seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [j0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 0 - i = ~ i; (s)% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)% ⊦ i < 0 → (~2 < i ↔ i = ~1); (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: j0