What is the universe of a sub algebra generated by $ \{(a \wedge b)\vee(c \wedge b') : a,c \in C\}$ ?

Question

I need to prove the next thing,

Let $B$ be a Boolean algebra and $C$ a proper subalgebra of $B$. Let $b ∈ B−C$. Prove that the set $ \{(a \wedge b)\vee(c \wedge b') : a,c \in C\}$ is the universe of a subalgebra of $B$ generated by $C ∪\{b\}$.

Now, I tried doing some things to prove the closure under join, meets and complements, but I am missing something, I would very much appreciate any hint on how to define things at first so I can do that part properly. Thank you very much.

Answer 1

Let $X =\{(a \wedge b) \vee (c \wedge b') \mid a, c \in C\}, Y$ is the universe of a subalgebra of $B$ generated by $Z = C \cup \{b\}.$ We want to show that $X = Y$.

1. $Y \subseteq X$: since $Y$ is the smallest subalgebra of $B$ containing $Z$ it is sufficient to show that $X$ is a subalgebra and $Z \subseteq X$. Fix some $c \in C: c = c \vee 1 = c \vee (b \wedge b') = (c \wedge b) \vee (c \wedge b')$. Hence $C \subseteq X$. Also $b = (1 \wedge b) \vee (b' \wedge 0)$. Hence $b \in X$, so $Z \subseteq X$. It is left to show that $X$ is the subalgebra of $B$. Fix arbitrary elements $x, y \in X:$ $$x = (a_x \wedge b) \vee (c_x \wedge b'),\quad y = (a_y \wedge b) \vee (c_y \wedge b').$$ Use the laws of a Boolean algebra to show that the following equalities hold: $$x \vee y = ((a_x \vee a_y) \wedge b) \vee ((c_x \vee c_y) \wedge b'),$$ $$x \wedge y = ((a_x \wedge a_y) \wedge b) \vee ((c_x \wedge c_y) \wedge b'),$$ $$x'= (a_x' \wedge b) \vee ((a_x' \vee c_x) \wedge b').$$ Since $C$ is a subalgebra of $B$ we have $x \wedge y, x \vee y, x' \in X$ hence $X$ is a subalgebra of $B$.
2. $X \subseteq Y$: since $Y$ is the subalgebra generated by $Z$, any Boolean term over a set of variables $Z$ is also in $Y$, but every element of $X$ is of the form $(a \wedge b) \vee (c \wedge b'),$ where $a, b, c \in Z$.