Help with getting the right direction on a boolean algebra question

Question

Need some help getting in the right direction for answering the following question:

Prove the following property and interpret this in $\mathcal P \left ({V} \right)$:

if $x+ \bar y=$ 1, then $x+y=x$.

How can I best approach this?

Answer 1

We start with $x$, then use the boolean axioms and the identity $x+\overline{y}=1$ to reach $x+y$. Here's a proof (I'll leave off the brackets, due to associativity): \begin{align*} x &= x+x & \text{idempotence of } + \\ &= x+x*1 & \text{identity of } * \\ &= x+x*(y+\overline{y}) & \text{complementation of } + \\ &= x+x*y+x*\overline{y} & \text{distributivity of } * \text{ over } + \\ &= x*x+x*y+x*\overline{y} & \text{idempotence of } * \\ &= x*x+x*y+x*\overline{y}+0 & \text{identity of } + \\ &= x*x+x*y+x*\overline{y}+y*\overline{y} & \text{complementation of } * \\ &= x*(x+y)+x*\overline{y}+y*\overline{y} & \text{distributivity of } * \text{ over } + \\ &= x*(x+y)+(x+y)*\overline{y} & \text{distributivity of } * \text{ over } + \\ &= x*(x+y)+\overline{y}*(x+y) & \text{commutativity of } * \\ &= (x+\overline{y})*(x+y) & \text{distributivity of } * \text{ over } + \\ &= 1*(x+y) & \text{by assumption } x+\overline{y}=1 \\ &= x+y & \text{identity of } *. \end{align*}