Question

Prove AU(BnC)=(AUB) n(AUC)​

Answer 1

Step-by-step explanation:

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Answer 2

Given:

A∪(B∩C)=(A∪B)∩(A∪C).

To Find:

Prove A∪(B∩C)=(A∪B)∩(A∪C).

Solution:

Here distributive property of sets is used in the expression, which states that For all sets  A,B  and  C ,  A∩(B∪C)=(A∩B)∪(A∩C)  and  A∪(B∩C)=(A∪B)∩(A∪C).

Here, in the given equation of different sets A, B, and C we can see that the Union U gets distributed over the intersection ∩.

If B intersects C and x is in A union, x is either in A or in (B and C). As a result, we must think about two scenarios:

• If x is present in A, x is also present in (A union B) and in A. (A union C). Consequently, x is where (A union B) intersects (A union C).
• If x is in both B and C, then x is also in both A union B and A union C because x is in both B and C. Therefore, x is once more in (A union B) intersect (A union C). This demonstrates: A∪(B∩C)⊂(A∪B)∩(A∪C).

The reverse inequality must be established in order to complete the proof. Consider x where (A union B) intersects (A union C). Consequently, x is both in (A or B) and in (A or C).

• x is also a member of the union of A if x is in A. (B intersect C).
• x must also be in C if it is in B. As a result, x is in B intersect C and is thus in A union (B intersect C). Thus, it can be seen that A∪(B∩C)⊃(A∪B)∩(A∪C)

Hence, both inequalities together demonstrate the equality of the two sets.

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