Possible Duplicate:
Cross Product of Partial Orders
Suppose that $(L_1;≤_1)$ and $(L_2;≤_2)$ are partially ordered sets. We define a partial order $≤$ on the set $L_1 \times L_2$ in the most obvious way - we say $(a,b)≤(c,d)$ if and only if $a≤_1 c$ and $b≤_2 d$.
a) Show that if $(L_1;≤_1)$ and $(L_2;≤_2)$ are both lattices, then so is $(L_1 \times L_2;≤)$.
b) Show that if $(L_1;≤_1)$ and $(L_2;≤_2)$ are both modular lattices, then so is $(L_1 \times L_2;≤)$.
c) Show that if $(L_1;≤_1)$ and $(L_2;≤_2)$ are both distributive lattices, then so is $(L_1 \times L_2;≤)$.
d) Show that if $(L_1;≤_1)$ and $(L_2;≤_2)$ are both Boolean algebras, then so is $(L_1 \times L_2;≤)$.
I am having trouble solving this problem. Any help would be appreciated.
I've been trying to attempt this solution. Here is where I get stuck:
c) $L_1$ and $L_2$ are modular, so
$$L_1: a_1 < a_3 \rightarrow a_1 \vee (a_2 \wedge a_3) = (a_1 \vee a_2) \wedge {a_3}$$
$$L_2: b_1 < b_3 \rightarrow b_1 \vee (b_2 \wedge b_3) = (b_1 \vee b_2) \wedge {b_3}$$
$$L_1\times L_2: (a_1,b_1) < (a_3,b_3) \rightarrow (a_1,b_1) \vee ((a_2,b_2) \wedge (a_3,b_3)) = ((a_1,b_1) \vee (a_2,b_2)) \wedge (a_3,b_3)$$
For $L_1\times L_2$ we have $$(a_1,b_1) < (a_3,b_3) = a_1 < a_3 \text{ and } b_1 < b_3$$
I'm not too sure how to use all this information to prove that $L_1\times L_2$ is a modular lattice. Any guidance would be greatly appreciated.