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-&gt;xAz 
S-&gt;BAx
A-&gt;Ay
A-&gt;e
B-&gt;yB
B-&gt;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($) , S-&gt;.BAx($) ) =

S-&gt;.xAz($)
S-&gt;.BAx($)
B-&gt;.yB(x,y)
B-&gt;.y(x,y)


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

S-&gt;x.Az($)
A-&gt;.Ay(z)
A-&gt;. (z) 


To find the lookahead symbol for A->. &amp; 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 $z$.