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Proof step 13 from proof div2-nls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
i + j = i0 * 2, j < i0, i ≤ j, i0 ≤ 0, j < 0, i < i0, i < 0 ⊦

Rule argument

Additional information

Used simplifier rules: zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < 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%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%numlsledel: j0 ≠ 0, j < j0 ⊦ i ≤ (* j) → i < (* j0); from specification int-abs%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs