Show that the direct product of Boolean algebra is again a Boolean algebra.
I defined the direct product of $B_1$ and $B_2$ as $(B_1\times B_2,\vee_3,\wedge_3,''',0_3,1_3)$ where the operations are defined for any $(a_1,b_1), (a_2,b_2)\in B_1\times B_2$ by
\begin{align*}
(a_1,b_1)\vee_3(a_2,b_2) &=(a_1\vee_1 a_2, b_1\vee_2 b_2)\\
(a_1,b_1)\wedge_3(a_2,b_2) &=(a_1\wedge_1 a_2, b_1\wedge_2 b_2)\\
(a_1,b_1)''' &=(a_1',b_1'')\\
0_3=(0_1,0_2)\quad &\text{and}\quad 1_3=(1_1,1_2)
\end{align*}
I am not able to show that $B_1\times B_2$ is a distributive lattice.
My try:
\begin{align*}
(a_1,b_1)\vee_3 \{(a_2,b_2)\wedge_3 (a_3,b_3)\} &= (a_1,b_1)\vee_3\{(a_2\wedge_1 a_3),(b_2\wedge_2 b_3)\}\\
&= \{a_1\vee_1 (a_2\wedge_1 a_3),b_1\vee_2 (b_2\wedge_2 b_3)\}\\
&= \{(a_1\vee_1 a_2)\wedge_1(a_1\vee_1 a_3),(b_1\vee_2 b_2)\wedge_2 (b_1\vee_2 b_3)\}
\end{align*}
Thanks for any help.
Best Answer
From the point you stand,
\begin{align} &= (a_1 \vee_1 a_2, b_1 \vee_2 b_2) \wedge_3 (a_1 \vee_1 a_3, b_1 \vee_2 b_3)\\ &= ((a_1, b_1) \vee_3 (a_2, b_2)) \wedge_3 ((a_1, b_1) \vee_3 (a_3, b_3)), \end{align} qed.