Let K be a field, with additive identity 0 and multiplicative identity 1. Let N be the set of natural numbers with successor map S as in that definition, ordered by the relations of Order on the Natural Numbers, let n∈N, and let [n] be the initial segment determined by n. Let a:[n]→K and b:[n]→K be maps, with values written ak and bk. All products below are the finite products of that definition.
Two different order relations occur below. In index ranges such as 1≤k≤n the symbol ≤ is the order on N just cited; in claim 5 it is the order of the ordered field of real numbers. Then the following hold.
1. (Restriction and recursion) If j∈[n] and a′ denotes the restriction of a to [j], then
k=1∏iak′=k=1∏iakfor every i∈[j].
Moreover
k=1∏1ak=a1,k=1∏S(m)ak=(k=1∏mak)aS(m)whenever S(m)∈[n].
2. (Multiplicativity)
k=1∏n(akbk)=(k=1∏nak)(k=1∏nbk).
3. (A single factor different from the unit) Let i∈[n] and suppose that ak=1 for every k∈[n] with k=i. Then
k=1∏nak=ai.
4. (Vanishing) ∏k=1nak=0 if and only if ai=0 for some i∈[n].
5. (Nonnegative factors) Suppose K is the field of real numbers, with the order ≤ of its ordered field structure, and that 0≤ak for every k∈[n]. Then
0≤k=1∏nak;
and if in addition ak≤bk for every k∈[n], then
k=1∏nak≤k=1∏nbk.