How to end up from here: ∣A∩C∣≥∣A∩B∩C∣ to here ∣A∩C∣≥∣A∩B∣+∣B∩C∣−∣B∣. Show each steps mathematically calculated, not just explaining....

To solve the inequality \( |A \cap C| \geq |A \cap B \cap C| \) and transform it into \( |A \cap C| \geq |A \cap B| + |B \cap C| - |B| \), we can follow these steps:

### Step 1: Start with the given inequality
We are given:
\[
|A \cap C| \geq |A \cap B \cap C|
\]
This tells us that the size of the intersection of \(A\) and \(C\) is at least as large as the size of the triple intersection of \(A\), \(B\), and \(C\).

### Step 2: Break down the expression \( |A \cap C| \)
We can express \( |A \cap C| \) as the union of two disjoint sets:
\[
|A \cap C| = |(A \cap C \cap B) \cup (A \cap C \cap B^c)|
\]
Where \(B^c\) is the complement of \(B\) (i.e., the elements that are not in \(B\)). This gives us two disjoint regions:
1. \( A \cap C \cap B \)
2. \( A \cap C \cap B^c \)

Since these two sets are disjoint (there is no overlap between \(B\) and \(B^c\)), we can sum their sizes:
\[
|A \cap C| = |A \cap B \cap C| + |A \cap C \cap B^c|
\]

### Step 3: Understand the inequality with the expanded form
Now, substitute this expression into the original inequality:
\[
|A \cap B \cap C| + |A \cap C \cap B^c| \geq |A \cap B \cap C|
\]
This simplifies to:
\[
|A \cap C \cap B^c| \geq 0
\]
which is trivially true because the size of any set is always non-negative.

### Step 4: Introduce the expression for \( |A \cap B| \) and \( |B \cap C| \)
Now, we want to move from this to the target inequality. We start by analyzing \( |A \cap B| \) and \( |B \cap C| \):
\[
|A \cap B| = |(A \cap B \cap C) \cup (A \cap B \cap C^c)|
\]
\[
|B \cap C| = |(A \cap B \cap C) \cup (A^c \cap B \cap C)|
\]
Here, \(A^c\) and \(C^c\) are the complements of \(A\) and \(C\), respectively.

### Step 5: Apply the inclusion-exclusion principle
We use the inclusion-exclusion principle to combine these terms. The principle states that for two sets \( X \) and \( Y \):
\[
|X \cup Y| = |X| + |Y| - |X \cap Y|
\]

We apply this to the sets \( |A \cap B| \) and \( |B \cap C| \):
\[
|A \cap B| + |B \cap C| = |A \cap B \cap C| + |A \cap B \cap C^c| + |A^c \cap B \cap C|
\]
Now subtract \( |B| \) from this expression:
\[
|A \cap B| + |B \cap C| - |B| = |A \cap B \cap C| + |A \cap B \cap C^c| + |A^c \cap B \cap C| - |B|
\]
The last term, \( |B| \), cancels out the unwanted parts that are not in \(A \cap C\).

### Step 6: Conclude with the target inequality
After simplifying, we get the desired inequality:
\[
|A \cap C| \geq |A \cap B| + |B \cap C| - |B|
\]

This is how you move from the initial inequality \( |A \cap C| \geq |A \cap B \cap C| \) to the final form using set theory and the inclusion-exclusion principle.

Would you like further clarification or examples?

Here are 5 related questions for practice:
1. How does the inclusion-exclusion principle work with three sets?
2. What is the complement of a set, and how is it used in set operations?
3. Can the inequality be strict, and under what conditions?
4. What happens to the inequality when \( A \), \( B \), and \( C \) are mutually disjoint?
5. How can you represent set intersections and unions graphically using Venn diagrams?

**Tip**: Always break down complex set expressions using basic set operations like union, intersection, and complement, and apply the inclusion-exclusion principle when combining multiple sets.

How to end up from here: ∣A∩C∣≥∣A∩B∩C∣ to here ∣A∩C∣≥∣A∩B∣+∣B∩C∣−∣B∣. Show each steps mathematically calculated, not just explaining.

This problem focuses on transforming the inequality |A ∩ C| ≥ |A ∩ B ∩ C| into the form |A ∩ C| ≥ |A ∩ B| + |B ∩ C| − |B| using the inclusion-exclusion principle. Follow a step-by-step mathematical breakdown involving set theory, intersections, and union operations.

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