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Proof step 2 from proof forward-pos (previous step)

i < j, 0 < i ⊦ 0 < j ∧ 1 < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 < j, i < j, 0 < i ⊦ 1 ≠ j ∧ ¬ j < 1

Rule argument

Additional information

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