I would like to apply an element chasing style proof to the following problem:
Let $P$, $Q$ and $R$ be elements of a Boolean algebra. Prove that if $ \tag 1 P + Q = P + R$ and $\tag 2 P \cdot Q = P \cdot R $ then $\tag 3 Q = R$
Given (1) above we know the $F(P,Q) = P + Q$ and $F(P,R) = P + R$ are both boolean functions of degree 2. Let $B = \{1,0\}$, which is the set of all Boolean variables. Then if we let $P \in B$ be arbitrary, we know that (1) may hold where $Q \in B $ is arbitrary and where $R \in B$ is arbitrary. That is (1) holds where $Q$ and $R$ are the same elements of the set $B$ and where they are different elements of $B$.
However, given that we are told (2) above holds and if we let $P \in B$ be arbitrary, we know that by the Identity ($P\cdot 1=P$) and Domination laws ($P\cdot 0 = 0$) that (2) can only hold where $Q$ and $R$ are the same $\in$ in $B$ and not when they are different elements of the set $B$.
$\therefore$ (3) must hold which states $Q = R$
Please could I get some feedback on this proof. Is it okay to apply an element chasing style to a problem like this or are other ways preferred? Could someone comment on whether the proof above goes far enough to conclusively prove that Q = R given (1) and (2)? thanks
Thanks