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Proof step 5 from proof coef-fac-divides (previous step)

0 < i, 0 < j, j ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → j ≤ i0 ∧ 0 < j ∧ 0 < i0 → fac(j) * fac(i0 - j) | fac(i0) ⊦ i = j, fac(j) * fac(i - j) | fac(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < j, j < i, 1 < i, 0 < i, ∀ i0, j. i→n(i0) < i→n(i) → j ≤ i0 ∧ 0 < j ∧ 0 < i0 → fac(j) * fac(i0 - j) | fac(i0) ⊦ i = j, fac(j) * fac(i - j) | fac(i)

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs