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Proof step 3 from proof le-pos (previous step)

2 ≤ i ∧ 0 ≤ j ∧ 1 < i, ∀ j1, i, j0. i→n(j1) < i→n(j) → 2 ≤ i ∧ 0 ≤ j1 ∧ 1 < i → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ i ^ j ≤ i ^ j0 ↔ j ≤ j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information