# LR(1) parsing, problem with look ahead symbols

I understand the concept of LR(1) parsing and lookahead symbols. I have the solution to the exercise and it does not agree with my solution.

I'm trying to fill the LR(1) parsing table for the grammar below:

S->xAz
S->BAx
A->Ay
A->e
B->yB
B->y


Ι don't have to extend the grammar since S does not appear in any right hand side of any rule.

First(A)=y,e
First(Ax)=x,y
First(B)=y
First(Ay)=y


Lookahead symbols in brackets.

So, I0 = Closure(S->.xAz($$) , S->.BAx($$) ) =

S->.xAz($$) S->.BAx($$)
B->.yB(x,y)
B->.y(x,y)


When i try GOTO(0,x) i think that i should go to:

S->x.Az(\$)
A->.Ay(z)
A->. (z)


To find the lookahead symbol for A->. & A->.Ay i take First(z). But the official book solution says the lookeahead is (z,y). Where does that y comes from?

Thank you in advance!

Since $$A\to .Ay$$ is in the state constructed for $$GOTO(0, x)$$, $$A\to .A(y)$$ and $$A\to .(y)$$ are also in that state. These combine with the other items for the same productions, which have lookahead $$z$$.