Equivalence of expressions involving conditional probabilities

Take the following random variables $$Y,\epsilon_0,\epsilon_1,V_0,V_1$$, with finite supports $$\mathcal{Y}\equiv \{0,1\},\mathcal{E}_0, \mathcal{E}_1,\mathcal{V}_0,\mathcal{V}_1$$, respectively.

Suppose $$P_Y(1|e_0,e_1,v_0,v_1)=\mathbb{1}\{e_0+v_0\leq e_1+v_1\}$$ $$\forall e_0\in \mathcal{E}_0$$,$$\forall e_1\in \mathcal{E}_1$$,$$\forall v_0\in \mathcal{V}_0$$,$$\forall v_1\in \mathcal{V}_1$$, where $$\mathbb{1}\{a\leq 0\}$$ is $$1$$ if $$a\leq 0$$ and $$0$$ otherwise.

I want to derive the unconditional probability distribution of $$Y$$.

I can use the law of total probability, that is $$(\diamond) \text{ }\text{ }\text{ }P_Y(1)=\sum_{e_0,e_1,v_1,v_0} \mathbb{1}\{e_0+v_0\leq e_1+v_1\} \times P_{\epsilon_0,\epsilon_1,V_0,V_1}$$

Question: Let $$W_0\equiv \epsilon_0+V_0$$ and $$W_1\equiv \epsilon_1+V_1$$. My intuition without going through the law of total probability is that $$(\diamond \diamond) \text{ }\text{ }\text{ }P_Y(1)= \sum_{w_0,w_1} \mathbb{1}\{w_0\leq w_1\}\times P_{W_0, W_1}(w_0, w_1)$$ In other words, knowing the joint distribution of the sums, $$P_{W_0, W_1}$$ (rather than the joint distribution of each term, $$P_{\epsilon_0,\epsilon_1,V_0,V_1}$$) is sufficient. Is this correct? If yes, how do we show that $$(\diamond)=(\diamond \diamond)$$?

Yes, it is sufficient. Letting $$w_0=e_0+v_0$$ and $$w_1=e_1+v_1$$, you can safely add functions of your RVs to the given side of the probability formulas, and then rename them: \begin{align}P_Y(1|e_0,e_1,v_0,v_1)&=P_Y(1|e_0,e_1,v_0,v_1,e_0+v_0,e_1+v_1)\\&=P_Y(1|e_0,v_0,e_1,v_1,w_0,w_1)\\&=\mathbb{1}\{w_0\leq w_1\}\end{align} Since, the RHS probability doesn't depend further on $$e_i,v_i$$, we can safely omit them in LHS: $$P_Y(1|w_0,w_1)=\mathbb{1}\{w_0\leq w_1\}$$ Now, you can rewrite your second formula, i.e. $$\diamond\diamond$$, as follows by changing the first multiplicand inside the summation expression: $$P_Y(1)=\sum_{w_0,w_1}P_Y(1|w_0,w_1)\times P_{W_0,W_1}(w_0,w_1)$$ this is always true due to Total Probability Law, which also applies to $$(\diamond)$$ saying also that these two are equivalent because they are both different versions of $$P_Y(1)$$.