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Proof step 1 from proof abs-ls-02 (conclusion)

⊦ j ≤ j0 ∧ j0 ≤ i → abs(j0 - j) ≤ abs(i - j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
abs(i - j) < abs(j0 - j) ⊦ j0 < j, i < j0, i < j

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;%lf-08: ⊦ ¬ i < j ∧ ¬ i0 < i → ¬ i0 < j; (forward) from specification int-basic1