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Proof step 1 from proof abs-sub-01 (conclusion)

⊦ i + 1 ≠ j ∧ i < j → ¬ abs(j - i) ≤ abs(j - (i + 1) + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i < j ⊦ i + 1 = j, abs(j - (i + 1) + 1) < abs(j - i)

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;