Probability $P(Y_1 = 2 | Y_2 = 0)$ where $Y_1 = X_1 + X_2$ and $Y_2 = X_1 - X_2$ - $X1, \, X_2 \in \lbrace 1,2,3,4 \rbrace$ with uniform distribution

I'd like to know how to evaluate the probability $P(Y_1 = 2 | Y_2 = 0)$ where $Y_1 = X_1 + X_2$ and $Y_2 = X_1 - X_2$, where $X1, \, X_2 \in \lbrace 1,2,3,4 \rbrace$ have a uniform distribution so $P(X_1=k)=P(X_2=k) = 1/4$ (where $k \in \lbrace 1,2,3,4 \rbrace$).

I can write:

$$ P(Y_1 = 2 | Y_2 = 0) = P(Y_1 = 2 | X_1 = 1, X_2 = 1) = ? $$

EDIT: I'm interested in the case $X_1=X_2=1$; $X_1$ and $X_2$ are independent. I asked the question in the wrong way. The goal is to evaluate the following probability:

$$ P(Y_1 = 2 | X_1 = 1, X_2 = 1) $$

Solution 1:

$$ \small\begin{align} \mathsf P(Y_1=2\mid Y_2=0)&=\mathsf P(X_1+X_2=2\mid X_1=X_2)&&\text{definition of }Y_1, Y_2\\[2ex]&=\dfrac{\mathsf P(X_1+X_2=2, X_1=X_2)}{\mathsf P(X_1=X_2)}&&\text{conditional probability}\\[2ex]&=\dfrac{\mathsf P(X_1=1,X_2=1)}{\sum_{k=1}^4\mathsf P(X_1=k,X_2=k)}&&\text{law of total probability}\\[2ex]&=\dfrac{1}{4}&&\text{independent & uniform discrete distributions} \end{align} $$