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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)
System applied rule normalize
And obtained the following new goals
Proof step 2
:
i < j, abs(j - i) ≤ abs(j - (i + 1) + 1) ⊦ i + 1 = j
Rule argument
Additional information