1. **Stating the problem:** We are given classes and relations with specific properties and asked: Given that $M$ and $N$ belong to $\mathrm{Sel}(P,A)$, $M \neq N$, and condition (A) holds, can $\mathrm{Sumrel}(D(M))$ be equal to $\mathrm{Sumrel}(D(N))$ under special cases? 2. **Recall definitions and conditions:** - $\mathrm{Sel}(P,A)$ contains relations $Q$ such that the inverse of $Q$ is a function with domain $A$ and $Q \subseteq P$. - $D(M)$ is the domain of relation $M$. - $\mathrm{Sumrel}(D(M))$ is the relation subsisting between $x$ and $y$ whenever for some relation $Q$ in $D(M)$, $xQy$. - Condition (A): For any distinct $z,w \in A$, $\mathrm{Sumrel}(\mathrm{Ref}(P,z))$ and $\mathrm{Sumrel}(\mathrm{Ref}(P,w))$ have no couples in common. 3. **Analyze the problem:** - Since $M,N \in \mathrm{Sel}(P,A)$, by definition, the inverse of $M$ and $N$ are functions with domain $A$. - $M \neq N$ means there exists some $a \in A$ such that $M^{-1}(a) \neq N^{-1}(a)$. - $\mathrm{Ref}(P,z)$ is the class of all $y$ such that $y P z$. - $\mathrm{Sumrel}(\mathrm{Ref}(P,z))$ collects all pairs $x,y$ such that for some $Q$ in $\mathrm{Ref}(P,z)$, $x Q y$. 4. **Use condition (A):** - For distinct $z,w \in A$, $\mathrm{Sumrel}(\mathrm{Ref}(P,z))$ and $\mathrm{Sumrel}(\mathrm{Ref}(P,w))$ are disjoint. - Since $M^{-1}$ and $N^{-1}$ are functions with domain $A$, for each $a \in A$, $M^{-1}(a)$ and $N^{-1}(a)$ are elements related to $a$ by $P$. 5. **Implication for $\mathrm{Sumrel}(D(M))$ and $\mathrm{Sumrel}(D(N))$:** - $D(M)$ and $D(N)$ are the domains of $M$ and $N$ respectively. - Since $M$ and $N$ are relations with inverses that are functions on $A$, their domains $D(M)$ and $D(N)$ correspond to the images of $M^{-1}$ and $N^{-1}$. - Because $M \neq N$, there exists $a \in A$ such that $M^{-1}(a) \neq N^{-1}(a)$. - By condition (A), the relations $\mathrm{Sumrel}(\mathrm{Ref}(P,M^{-1}(a)))$ and $\mathrm{Sumrel}(\mathrm{Ref}(P,N^{-1}(a)))$ have no couples in common. 6. **Conclusion:** - Since $\mathrm{Sumrel}(D(M))$ includes $\mathrm{Sumrel}(\mathrm{Ref}(P,M^{-1}(a)))$ and $\mathrm{Sumrel}(D(N))$ includes $\mathrm{Sumrel}(\mathrm{Ref}(P,N^{-1}(a)))$, and these are disjoint for distinct elements by (A), it follows that $\mathrm{Sumrel}(D(M))$ and $\mathrm{Sumrel}(D(N))$ cannot be equal. **Final answer:** $$\boxed{\text{No, } \mathrm{Sumrel}(D(M)) \neq \mathrm{Sumrel}(D(N)) \text{ under the given conditions.}}$$