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Proof step 1 from proof l (conclusion)

⊦ ¬ i < j ↔ ¬ ¬(i = j ∨ j < i)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ls-01: ⊦ i < j → ¬ j < i;%lst-02: ⊦ ¬ i < j → (¬ j < i ↔ i = j);