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Proof step 9 from proof div2-nls-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
i0 * 2 = (i + j) + 1, i0 < i, i ≤ j, i + j < 0, i0 ≤ 0, i0 < j ⊦

Rule argument

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; 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%minus-add: i < 0 ⊦ (* i) + j = j - ~ i; from specification int-abs