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Proof step 5 from proof abs-neg (previous step)

0 = i0 + i ⊦ abs(i0) = abs(i), i0 < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: Abs-pos: ⊦ ¬ j < 0 → abs(j) = j;% ⊦ i + j < j ↔ i < 0; (s)%lf-05: ⊦ ¬ j < i ∧ j = i0 → ¬ i0 < i; (forward) from specification int-basic1%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% ⊦ i < i + j ↔ 0 < j; (s)%lf-06: ⊦ ¬ j < i ∧ i = i0 → ¬ j < i0; (forward) from specification int-basic1