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Proof step 2 from proof numlsf-01 (previous step)

i < j, 0 < i ⊦ 0 ≤ j ∧ 0 < j

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i; (s)% ⊦ i < j → ¬ j < i; (s)