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Sagot :
Sure, let's work through the given set identities step by step.
### First Property: [tex]\( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)[/tex]
To prove this, we need to show that an element [tex]\( x \)[/tex] belongs to the set on the left-hand side if and only if it belongs to the set on the right-hand side.
Proof:
1. Suppose [tex]\( x \in A \cap (B \cup C) \)[/tex].
- By the definition of intersection, [tex]\( x \in A \)[/tex] and [tex]\( x \in B \cup C \)[/tex].
- By the definition of union, [tex]\( x \in B \cup C \)[/tex] means that [tex]\( x \in B \)[/tex] or [tex]\( x \in C \)[/tex].
Therefore, we have two cases:
- If [tex]\( x \in B \)[/tex], then [tex]\( x \in A \cap B \)[/tex].
- If [tex]\( x \in C \)[/tex], then [tex]\( x \in A \cap C \)[/tex].
Thus, [tex]\( x \in (A \cap B) \cup (A \cap C) \)[/tex].
2. Conversely, suppose [tex]\( x \in (A \cap B) \cup (A \cap C) \)[/tex].
- By the definition of union, [tex]\( x \in (A \cap B) \cup (A \cap C) \)[/tex] means that [tex]\( x \in A \cap B \)[/tex] or [tex]\( x \in A \cap C \)[/tex].
Therefore, we have two cases:
- If [tex]\( x \in A \cap B \)[/tex], then [tex]\( x \in A \)[/tex] and [tex]\( x \in B \)[/tex].
- If [tex]\( x \in A \cap C \)[/tex], then [tex]\( x \in A \)[/tex] and [tex]\( x \in C \)[/tex].
- In either case, we have [tex]\( x \in A \)[/tex] and [tex]\( x \in B \cup C \)[/tex].
Thus, [tex]\( x \in A \cap (B \cup C) \)[/tex].
Since both directions have been shown, we conclude that [tex]\( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)[/tex].
### Second Property: [tex]\( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \)[/tex]
To prove this, we need to show that an element [tex]\( x \)[/tex] belongs to the set on the left-hand side if and only if it belongs to the set on the right-hand side.
Proof:
1. Suppose [tex]\( x \in A \cup (B \cap C) \)[/tex].
- By the definition of union, [tex]\( x \in A \cup (B \cap C) \)[/tex] means that [tex]\( x \in A \)[/tex] or [tex]\( x \in B \cap C \)[/tex].
- If [tex]\( x \in B \cap C \)[/tex], then [tex]\( x \in B \)[/tex] and [tex]\( x \in C \)[/tex].
Therefore, we have two cases:
- If [tex]\( x \in A \)[/tex], then [tex]\( x \in A \cup B \)[/tex] and [tex]\( x \in A \cup C \)[/tex].
- If [tex]\( x \in B \cap C \)[/tex], then [tex]\( x \in B \)[/tex] and [tex]\( x \in C \)[/tex], implying that [tex]\( x \in A \cup B \)[/tex] and [tex]\( x \in A \cup C \)[/tex].
Thus, [tex]\( x \in (A \cup B) \cap (A \cup C) \)[/tex].
2. Conversely, suppose [tex]\( x \in (A \cup B) \cap (A \cup C) \)[/tex].
- By the definition of intersection, [tex]\( x \in (A \cup B) \cap (A \cup C) \)[/tex] means that [tex]\( x \in A \cup B \)[/tex] and [tex]\( x \in A \cup C \)[/tex].
Therefore, we have two cases:
- If [tex]\( x \in A \)[/tex], then [tex]\( x \in A \)[/tex] which satisfies the left-hand side.
- If [tex]\( x \notin A \)[/tex], then [tex]\( x \in B \)[/tex] (as [tex]\( x \in A \cup B \)[/tex] and [tex]\( x \notin A \)[/tex]) and [tex]\( x \in C \)[/tex] (as [tex]\( x \in A \cup C \)[/tex] and [tex]\( x \notin A \)[/tex]), which implies [tex]\( x \in B \cap C \)[/tex].
In either case, we have [tex]\( x \in A \cup (B \cap C) \)[/tex].
Since both directions have been shown, we conclude [tex]\( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \)[/tex].
By showing both properties step by step with logical deductions, we have proven the set identities.
### First Property: [tex]\( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)[/tex]
To prove this, we need to show that an element [tex]\( x \)[/tex] belongs to the set on the left-hand side if and only if it belongs to the set on the right-hand side.
Proof:
1. Suppose [tex]\( x \in A \cap (B \cup C) \)[/tex].
- By the definition of intersection, [tex]\( x \in A \)[/tex] and [tex]\( x \in B \cup C \)[/tex].
- By the definition of union, [tex]\( x \in B \cup C \)[/tex] means that [tex]\( x \in B \)[/tex] or [tex]\( x \in C \)[/tex].
Therefore, we have two cases:
- If [tex]\( x \in B \)[/tex], then [tex]\( x \in A \cap B \)[/tex].
- If [tex]\( x \in C \)[/tex], then [tex]\( x \in A \cap C \)[/tex].
Thus, [tex]\( x \in (A \cap B) \cup (A \cap C) \)[/tex].
2. Conversely, suppose [tex]\( x \in (A \cap B) \cup (A \cap C) \)[/tex].
- By the definition of union, [tex]\( x \in (A \cap B) \cup (A \cap C) \)[/tex] means that [tex]\( x \in A \cap B \)[/tex] or [tex]\( x \in A \cap C \)[/tex].
Therefore, we have two cases:
- If [tex]\( x \in A \cap B \)[/tex], then [tex]\( x \in A \)[/tex] and [tex]\( x \in B \)[/tex].
- If [tex]\( x \in A \cap C \)[/tex], then [tex]\( x \in A \)[/tex] and [tex]\( x \in C \)[/tex].
- In either case, we have [tex]\( x \in A \)[/tex] and [tex]\( x \in B \cup C \)[/tex].
Thus, [tex]\( x \in A \cap (B \cup C) \)[/tex].
Since both directions have been shown, we conclude that [tex]\( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)[/tex].
### Second Property: [tex]\( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \)[/tex]
To prove this, we need to show that an element [tex]\( x \)[/tex] belongs to the set on the left-hand side if and only if it belongs to the set on the right-hand side.
Proof:
1. Suppose [tex]\( x \in A \cup (B \cap C) \)[/tex].
- By the definition of union, [tex]\( x \in A \cup (B \cap C) \)[/tex] means that [tex]\( x \in A \)[/tex] or [tex]\( x \in B \cap C \)[/tex].
- If [tex]\( x \in B \cap C \)[/tex], then [tex]\( x \in B \)[/tex] and [tex]\( x \in C \)[/tex].
Therefore, we have two cases:
- If [tex]\( x \in A \)[/tex], then [tex]\( x \in A \cup B \)[/tex] and [tex]\( x \in A \cup C \)[/tex].
- If [tex]\( x \in B \cap C \)[/tex], then [tex]\( x \in B \)[/tex] and [tex]\( x \in C \)[/tex], implying that [tex]\( x \in A \cup B \)[/tex] and [tex]\( x \in A \cup C \)[/tex].
Thus, [tex]\( x \in (A \cup B) \cap (A \cup C) \)[/tex].
2. Conversely, suppose [tex]\( x \in (A \cup B) \cap (A \cup C) \)[/tex].
- By the definition of intersection, [tex]\( x \in (A \cup B) \cap (A \cup C) \)[/tex] means that [tex]\( x \in A \cup B \)[/tex] and [tex]\( x \in A \cup C \)[/tex].
Therefore, we have two cases:
- If [tex]\( x \in A \)[/tex], then [tex]\( x \in A \)[/tex] which satisfies the left-hand side.
- If [tex]\( x \notin A \)[/tex], then [tex]\( x \in B \)[/tex] (as [tex]\( x \in A \cup B \)[/tex] and [tex]\( x \notin A \)[/tex]) and [tex]\( x \in C \)[/tex] (as [tex]\( x \in A \cup C \)[/tex] and [tex]\( x \notin A \)[/tex]), which implies [tex]\( x \in B \cap C \)[/tex].
In either case, we have [tex]\( x \in A \cup (B \cap C) \)[/tex].
Since both directions have been shown, we conclude [tex]\( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \)[/tex].
By showing both properties step by step with logical deductions, we have proven the set identities.
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