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Proof step 1 from proof absp-01 (conclusion)

j ≤ j0, i ≤ j ⊦ abs(j - i) ≤ abs(j0 - i)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: abs-ls-02: ⊦ j ≤ j0 ∧ j0 ≤ i → abs(j0 - j) ≤ abs(i - j);