Fill in the Boolean Algebra axiom or theorem $($e$.$g$.,1$a$,1$b$,2$a$,$ etc.$)$ for each step of the following Boolean algebra minimization.

$(x+y)(x+!y)=+xy+x!y+y!y$

$(x+y)(x+y)=x+xy+x!y+y!y$

$(x+y)(x+!y)=x+xy+x!y+0$

$(x+y)(x+!y)=x+xy+x!y$

$(x+y)(x+!y)=x(1+y+!y)$

$(x+y)(x+!y)=x(1+1)$

$(x+y)(x+!y)=x(1)$

$(x+y)(x+!y)=x$