I do not want a proof! I'm just trying to understand the question since it seems false to me. The following is Exercise 7.10 from the Logic of Mathematics by Adamowicz and Zbierski:
Prove the generalized distributivity laws for Boolean algebras:
$$\sum_{i=1}^n a_{i1}\cdot \cdots \cdot a_{im} =
\prod_{j_1,\ldots,j_n=1}^m (a_{1j_1}+\cdots+a_{nj_n})$$
and
$$\prod_{i=1}^n (a_{i1}+\cdots + a_{im}) = \sum_{j_1,\ldots,j_n=1}^m a_{1j_1}\cdot \cdots \cdot a_{nj_n}$$
for an arbitrary matrix $\{a_{ij}: i=1,\ldots, n,j=1,\ldots,m\}$ of elements of the algebra.
But I do not understand the RHS. An example… if $a=[a_{11},a_{12};a_{21},a_{22}]$ (the 2×2 matrix) then the LHS of the first centered equation equals $a_{11}a_{12}+a_{21}a_{22}$. However, I do not know how to start with the RHS. Thanks!
Best Answer
The right-hand side just means a simultaneous product (or sum in the second case) where each $j_i$ independently ranges from $1$ to $n$. If you like, the symbol $$\prod_{j_1,\ldots,j_n=1}^m$$ means the same thing as an iterated product $$\prod_{j_1=1}^m\prod_{j_2=1}^m\dots\prod_{j_n=1}^m.$$
So for instance, when $n=m=2$, you get $(a_{11}+a_{21})(a_{11}+a_{22})(a_{12}+a_{21})(a_{12}+a_{22})$.
This should look more familiar in the case of the second distributive law, which is just the usual generalized distributive law for multiplication over addition (which holds in any ring, for instance). For instance, when $n=m=2$, it just says $$(a_{11}+a_{12})(a_{21}+a_{22})=a_{11}a_{21}+a_{11}a_{22}+a_{12}a_{21}+a_{12}a_{22}.$$ The first distributive law is just the same thing with the roles of addition and multiplication swapped.