Prove the following properties for any three sets A∩(B∪C)=(A∩B)∪(A∩C) andA∪(B∩C)=(A∪B)∩(A∪C)

Question

Prove the following properties for any three sets

A \cap (B \cup C) = (A \cap B) \cup (A \cap C)

and

A \cup (B \cap C) = (A \cup B) \cap (A \cup C)

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Solution

Sure, let's prove these properties one by one.

A∩(B∪C)=(A∩B)∪(A∩C)

Proof:

Let x be an arbitrary element of A∩(B∪C). Then x is in A and x is in B∪C. This means that x is in A and x is in B or x is in C. Therefore, x is in A∩B or x is in A∩C. So, x is in (A∩B)∪(A∩C). Hence, A∩(B∪C) is a subset of (A∩B)∪(A∩C).

Now, let x be an arbitrary element of (A∩B)∪(A∩C). Then x is in A∩B or x is in A∩C. This means that x is in A and x is in B or x is in A and x is in C. Therefore, x is in A and x is in B∪C. So, x is in A∩(B∪C). Hence, (A∩B)∪(A∩C) is a subset of A∩(B∪C).

Since A∩(B∪C) is a subset of (A∩B)∪(A∩C) and (A∩B)∪(A∩C) is a subset of A∩(B∪C), we conclude that A∩(B∪C)=(A∩B)∪(A∩C).

A∪(B∩C)=(A∪B)∩(A∪C)

Proof:

Let x be an arbitrary element of A∪(B∩C). Then x is in A or x is in B∩C. This means that x is in A or x is in B and x is in C. Therefore, x is in A or x is in B and x is in A or x is in C. So, x is in (A∪B) and x is in (A∪C). Hence, A∪(B∩C) is a subset of (A∪B)∩(A∪C).

Now, let x be an arbitrary element of (A∪B)∩(A∪C). Then x is in A∪B and x is in A∪C. This means that x is in A or x is in B and x is in A or x is in C. Therefore, x is in A or x is in B and x is in C. So, x is in A∪(B∩C). Hence, (A∪B)∩(A∪C) is a subset of A∪(B∩C).

Since A∪(B∩C) is a subset of (A∪B)∩(A∪C) and (A∪B)∩(A∪C) is a subset of A∪(B∩C), we conclude that A∪(B∩C)=(A∪B)∩(A∪C).

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Prove the following properties for any three sets A∩(B∪C)=(A∩B)∪(A∩C) andA∪(B∩C)=(A∪B)∩(A∪C)

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