Filter out information in Kalman Filter

I have a disagreement with friend over if an observable that depend on lagged observable, should be called a state variable. I have the following system:

xt=u+z(1,t-1)+x(t-1)+d·[y(t-1)-x(t-1)]+e(x,t)
yt=u+z(2,t-1)+y(t-1)-d·[y(t-1)-x(t-1)]+e(y,t)

z1 and z2 are AR(1) processes
z(1,t)=p1z(1,t-1)+e(1,t)
z(2,t)=p2z(2,t-1)+e(2,t)


The observations are xt and yt

I say the state variables are z1 and z2, but my friend say I am wrong in that the state variable must also include x(t-1) and y(t-1)

Since I think that x(t-1) and y(t-1) are observed and known at time t, can I claim that they are known and thus can be filtered out And as far as the Kalman filter gain and the variance of estimate P , as well as their steady state values are concerned, I can derive K and P and their steady state values by pretending the system is the following, which will yield the exact same K and P:

xt=u+z(1,t-1)+x(t-1)+e(x,t)
yt=u+z(2,t-1)+y(t-1)+e(y,t)

z1 and z2 are AR(1) processes
z(1,t)=p1z(1,t-1)+e(1,t)
z(2,t)=p2z(2,t-1)+e(2,t)


Am I right that the 2nd system will yield the same K and P and their respective steady state value as the first system?

But my friend say I am wrong because x(t-1) and y(t-1) are not known at time 0, they must be state variables, and thus xt and yt are both observables variables and state variables.

-
I do not understand your "d" symbol. If I was sure what it meant, then I could understand the AR part of your question and answer it. Why can you pretend it is zero with no consequence? –  EngrStudent Apr 15 at 4:54