How do I solve this Boolean Algebra Problem?

Question

Let A be an arbitrary but fixed Boolean algebra with operator $@$ and $*$ and $'$ and the zero and unit element be denoted by $0$ and $1$ respectively. let $x,y,z \in A$

if $a,y \in A$ such that $x*y=0$ and $x@y=1$ Then prove that $y=x'$

Answer 1

From the first condition you get:

$(x'*y) @ (x*y) = (x'*y) \Rightarrow (x'@x)*y = (x'*y) \Rightarrow y = x'*y$, and finally $y @ x' = (x'*y)@x' \Rightarrow y@x' = x'$.

Furthermore, with the second condition you get $(x'@y)*(x@y) = (x'@y) \Rightarrow (x'*x)@y = (x'@y) \Rightarrow y = x'@y $

Hence, $x' = y@x' = x'@y = y$, as required.