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Proof step 4 from proof abs-ls-02 (previous step)

i = i0 + j, abs(i0) < abs(j0 - j) ⊦ j0 < j, i < j0, i < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i0 < abs(j0 - j) ⊦ i0 + j < j0, j0 < j, i0 < 0

Rule argument

Additional information

Used simplifier rules: Abs-pos: ⊦ ¬ j < 0 → abs(j) = j;%lf-08: ⊦ ¬ i < j ∧ ¬ i0 < i → ¬ i0 < j; (forward) from specification int-basic1%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2