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Proof step 1 from proof lf-01 (conclusion)

⊦ i < j ∧ j < i0 → i +1 < i0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: l: ⊦ ¬ i < j ↔ ¬ ¬(i = j ∨ j < i);%ls-succ: ⊦ i < j +1 ↔ ¬ j < i;%trans: ⊦ i < j ∧ j < j0 → i < j0;