Sure! Let's break down the expression [tex]\((A \cup B) \cap (A \cup \bar{B})\)[/tex] to prove that it equals [tex]\(A\)[/tex].
### Step-by-Step Solution:
1. Expression Breakdown:
We start with the expression [tex]\((A \cup B) \cap (A \cup \bar{B})\)[/tex].
2. Distributive Property:
We will use the distributive property of sets, which states that:
[tex]\[ X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z) \][/tex]
3. Apply Distributive Property:
Applying the distributive property to our expression:
[tex]\[ (A \cup B) \cap (A \cup \bar{B}) = [(A \cup B) \cap A] \cup [(A \cup B) \cap \bar{B}] \][/tex]
4. Simplify Each Part:
We will now simplify each part individually:
[tex]\[ (A \cup B) \cap A \][/tex]
and
[tex]\[ (A \cup B) \cap \bar{B} \][/tex]
5. Simplify [tex]\( (A \cup B) \cap A \)[/tex]:
Intersection with [tex]\(A\)[/tex] distributes over the union:
[tex]\[ (A \cup B) \cap A = (A \cap A) \cup (B \cap A) = A \cup (A \cap B) \][/tex]
Since any set intersecting with itself is the set itself:
[tex]\[ A \cup (A \cap B) = A \][/tex]
6. Simplify [tex]\( (A \cup B) \cap \bar{B} \)[/tex]:
In a similar vein:
[tex]\[ (A \cup B) \cap \bar{B} = (A \cap \bar{B}) \cup (B \cap \bar{B}) \][/tex]
Since [tex]\(B \cap \bar{B} = \emptyset\)[/tex] (B and its complement do not overlap):
[tex]\[ (A \cap \bar{B}) \cup \emptyset = A \cap \bar{B} \][/tex]
7. Combine Results:
So we have:
[tex]\[ (A \cup B) \cap (A \cup \bar{B}) = [A] \cup [A \cap \bar{B}] \][/tex]
Since [tex]\(A \cap \bar{B} \subseteq A\)[/tex]:
[tex]\[ A \cup (A \cap \bar{B}) = A \][/tex]
### Conclusion:
Thus, we have shown that:
[tex]\[ (A \cup B) \cap (A \cup \bar{B}) = A \][/tex]
Hence, the expression [tex]\((A \cup B) \cap (A \cup \bar{B})\)[/tex] simplifies to [tex]\(A\)[/tex].
