Confusion about sum of random variables conditional probabilities

Let X,Y be independent random variables, and $Z=X+Y.$. Then I want to calculate $Pr[X=x\mid Z=z].$. My confusion is on evaluating this expression. On the one hand, I have

$Pr[X=x\mid Z=z]=Pr[Z-Y=x\mid Z=z]=Pr[z-Y=x]=Pr[Y=z-x].$

But also, $Pr[Z=z\mid X=x]=Pr[X+Y=z\mid X=x]=Pr[Y=z-x].$ So these two probabilities are equal? But $Pr[Z=z\mid X=x]Pr[X=x]=Pr[X=x\mid Z=z]Pr[Z=z]$ and in general $Pr[X=x]$ and $Pr[Z=z]$ are not equal. I believe it should be $Pr[X=x\mid Z=z]=Pr[Y=z-x\mid Z=z]$ but I don't think the conditional $Z=z$ can be removed since Y and Z are not independent?

I'm not sure whether the first equation holds either. For example, if I roll a fair six sided die X (numbered 1 to 6) and roll a fair ten sided die Y and take the sum, then $Pr[X=1\mid Z=2]=1$ since the only possible outcome is $(x,y)=(1,1),$, and this is not equal to $Pr[Y=(2-1)]=1/10.$. On the other hand it is equal to $Pr[Y=(2-1)\mid Z=2]=1.$. I think I'm making a mistake in one of these but it's not clear to me in which step.

(The context of this was that X is a random variable with given distribution representing some unknown parameter and Y is a standard normal error. Then you observe $z=x+y$ and want to estimate the X.)

Let X,Y be independent random variables, and $Z=X+Y.$. Then I want to calculate $Pr[X=x\mid Z=z].$. My confusion is on evaluating this expression. On the one hand, I have

$Pr[X=x\mid Z=z]=Pr[Z-Y=x\mid Z=z]=Pr[z-Y=x]=Pr[Y=z-x].$

But also, $Pr[Z=z\mid X=x]=Pr[X+Y=z\mid X=x]=Pr[Y=z-x].$ So these two probabilities are equal? But $Pr[Z=z\mid X=x]Pr[X=x]=Pr[X=x\mid Z=z]Pr[Z=z]$ and in general $Pr[X=x]$ and $Pr[Z=z]$ are not equal. I believe it should be $Pr[X=x\mid Z=z]=Pr[Y=z-x\mid Z=z]$ but I don't think the conditional $Z=z$ can be removed since Y and Z are not independent?

I'm not sure whether the first equation holds either. For example, if I roll a fair six sided die X (numbered 1 to 6) and roll a fair ten sided die Y and take the sum, then $Pr[X=1\mid Z=2]=1$ since the only possible outcome is $(x,y)=(1,1),$, and this is not equal to $Pr[Y=(2-1)]=1/10.$. On the other hand it is equal to $Pr[Y=(2-1)\mid Z=2]=1.$. I think I'm making a mistake in one of these but it's not clear to me in which step.

(The context of this was that X is a random variable with given distribution representing some unknown parameter and Y is a standard normal error. Then you observe $z=x+y$ and want to estimate the X.)