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

0 < i ∧ 0 < j, ∀ i0, j. i→n(i0) < i→n(i) → 0 < i0 ∧ 0 < j → (fac(i0) ≤ fac(j) ↔ i0 ≤ j) ⊦ fac(i) ≤ fac(j) ↔ i ≤ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information