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prove that A n (B u C) = (A n B) u ( A n C)​

Sagot :

Let x be inside set A n (B u C)

We can write this as [tex]x \in A \cap(B \cup C)[/tex]

This means x is in set A and also in set B u C. It may be in B, or in C, or both B and C at the same time.

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Let's assume x is in set B but not in set C.

If so, then x is definitely in A n B and not in A n C. Overall, x is in (A n B) u ( A n C)​

This leads us to the fact that A n (B u C) is a subset of (A n B) u ( A n C)​

Symbolically we would write [tex]A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)[/tex]

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Now let's assume that x is in set C but not in set B

We follow the same basic steps as the previous section. We'll conclude with the same symbolic notation written. There's not much different here other than x is in set A n C rather than A n B

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The previous two sections fully confirm that [tex]A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)[/tex] is true.

To prove [tex]A \cap (B \cup C) \supseteq (A \cap B) \cup (A \cap C)[/tex] is true, we will use the same idea.

Let y be a member of the set (A n B) u ( A n C)​

This means y is either in A n B or it is in A n C

If y is in A n B, then it's certainly in A and B. It doesn't matter if it's in C or not. We have enough to show that y is a member of  A n (B u C)

Or if y is in A n C, then it's in A and C simultaneously. This places element y in the set A n (B u C) overall.

Therefore, we have proven that [tex]A \cap (B \cup C) \supseteq (A \cap B) \cup (A \cap C)[/tex] is true. If something is in the right-side set, then it's also in the left-side set as well.

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To wrap things up, we have proven these two claims

[tex]A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)\\\\A \cap (B \cup C) \supseteq (A \cap B) \cup (A \cap C)[/tex]

In other words, we have shown the left and right sides are subsets of each other. This is sufficient to conclude that the two sets are equal.

Therefore, we have shown that

[tex]A \cap (B \cup C) = (A \cap B) \cup (A \cap C)[/tex]

is a true claim.

For more information, check out the distributive property dealing with set mathematics.

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