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Proof step 25 from proof coef-rec-add-01 (previous step)

0 < j, fac(i - 1) / (fac(i - 1 + j) * j * fac(j - 1)) = ((i - j) * fac(i - 1)) / (fac(i - j) * j * fac(j - 1)), fac(i - 1) / (fac(i - j) * fac(j - 1)) = (j * fac(i - 1)) / (fac(i - j) * j * fac(j - 1)), 0 ≤ j, 0 ≤ i, j ≤ i, j < i, 0 < i, 0 < fac(i - j) * j * fac(j - 1), 0 ≤ fac(i - 1), fac(j - 1) * fac(i - j) | fac(i - 1), 1 < j, 0 < j, 1 < i ⊦ fac(i) / (fac(i - j) * j * fac(j - 1)) = (j * fac(i - 1)) / (fac(i - j) * j * fac(j - 1)) + ((i - j) * fac(i - 1)) / (fac(i - j) * j * fac(j - 1)), (j * fac(i - 1)) % (fac(i - j) * j * fac(j - 1)) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: fac-not-zero: ⊦ fac(i) * fac(j) ≠ 0;%fac-zero: ⊦ i ≤ 0 → fac(i) = 1;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j); from specification int-mult%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%zero-eq: ⊦ i * j = 0 ↔ i = 0 ∨ j = 0; from specification int-mult%mod-mult-02: 0 ≠ j, i0 ≠ 0 ⊦ (i0 * i) % (i0 * j) = i % j * i0; from specification int-div%one: ⊦ 1 * i = i; from specification int-mult%div-mult-01: j ≠ 0, i0 ≠ 0 ⊦ (i * i0) / (j * i0) = i / j; from specification int-div%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%sub-zero-01: ⊦ i - 0 = i; from specification int-abs%Mult-zero: ⊦ i * 0 = 0; from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult