Show 
𝑀
∩
𝑃
≠
∅
M∩P

=∅, and 
𝑀
⊆
𝑆
M⊆S, checking if 
𝑆
∩
𝑃
S∩P matches.

You are asking whether the intersection of two sets \( M \) and \( P \), denoted \( M \cap P \), is non-empty, given that \( M \subseteq S \). Specifically, you want to check if \( S \cap P \) matches \( M \cap P \).

To analyze this, let's break down the steps:

### 1. Set Relationships:
- **\( M \cap P \neq \emptyset \)** means that sets \( M \) and \( P \) share at least one common element (i.e., their intersection is not empty).
- **\( M \subseteq S \)** means that every element of \( M \) is also an element of \( S \). This gives us the key information to analyze the intersection between \( S \) and \( P \).

### 2. The Intersection \( S \cap P \):
We are asked if the intersection of \( S \) and \( P \), denoted \( S \cap P \), "matches" \( M \cap P \). Here's the reasoning:

- Since \( M \subseteq S \), every element in \( M \) is also in \( S \). Thus, the intersection of \( M \) and \( P \) is also a subset of the intersection of \( S \) and \( P \). This can be written as:
  \[
  M \cap P \subseteq S \cap P.
  \]
  
- The question asks if \( M \cap P = S \cap P \). This would require that there are no elements in \( S \cap P \) that are not in \( M \cap P \).

### 3. Possible Scenarios:
- If \( M \cap P \) is non-empty, it means that there is at least one element common to both \( M \) and \( P \).
- Since \( M \subseteq S \), every element of \( M \) is also in \( S \). Therefore, \( M \cap P \) is a subset of \( S \cap P \), but it is not necessarily equal to \( S \cap P \). For \( M \cap P \) to be equal

Show M ∩ P ≠ ∅, and M ⊆ S, checking if S ∩ P matches.

Explore the concepts of set intersection and subset relations with the problem involving M ∩ P and S ∩ P. Understand how M ⊆ S influences the result of set intersections. Suitable for students studying set theory in Grades 10-12.

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