To solve part (в) of the problem, let's analyze the statement:

Statement:

$P_1^{-1} \cup P_2^{-1} = (P_1 \cup P_2)^{-1}$

Where:

1. $P_1^{-1}$ and $P_2^{-1}$ represent the inverse (transpose) of the binary relations $P_1$ and $P_2$.
2. $\cup$ denotes the union of sets.
3. The expression $(P_1 \cup P_2)^{-1}$ represents the inverse of the union of $P_1$ and $P_2$.

Proof:

Left-hand side (LHS):

The union of the inverses is defined as: $P_1^{-1} \cup P_2^{-1} = \{(b, a) \mid (a, b) \in P_1 \text{ or } (a, b) \in P_2\}.$

Right-hand side (RHS):

The inverse of the union is defined as: $(P_1 \cup P_2)^{-1} = \{(b, a) \mid (a, b) \in P_1 \cup P_2\}.$

Comparison:

• By definition of set union, $(a, b) \in P_1 \cup P_2$ if $(a, b) \in P_1$ or $(a, b) \in P_2$.
• Taking the inverse flips each pair, so both the LHS and RHS yield the same set: $P_1^{-1} \cup P_2^{-1} = (P_1 \cup P_2)^{-1}.$

Conclusion:

The statement in part (в) is true.

Let me know if you want a deeper explanation or further assistance with other parts.

Here are 5 questions for deeper understanding:

1. What is the significance of the inverse operation for binary relations?
2. How does the union operation interact with other set-theoretic operations?
3. Are there other logical operations where the distributive property might fail for relations?
4. Can you provide an example of $P_1$ and $P_2$ to verify the correctness of this equality?
5. How would the properties of the intersection differ from those of the union in this context?

Tip: Always verify these kinds of equivalences by testing with concrete examples of binary relations.