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Proof step 13 from proof fac-add (previous step)

1 ≤ j, 0 < i + j, 0 ≤ i, 0 < j, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ 0 ≤ j0 → fac(i + j0) = fac(i) * Π countFun(1, j0, λ j0. i + j0) ⊦ j = 0, (i + j) * fac(i + j - 1) = fac(i) * Π(countFun(1, j - 1, λ j0. i + j0) + (i + j)')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%eq-pos: ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%product-one: ⊦ Π i ' = i; from specification intlist%product-app: ⊦ Π(ints + ints0) = Π ints * Π ints0; from specification intlist