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# Let $A=\left\{ a,b,c,d,e,f,g,x,y,z \right\}$, $B=\left\{ 1,2,c,d,e \right\}$, $C=\left\{ d,e,f,g,2,y \right\}$. Verify $A\backslash \left( B\cup C \right)=\left( A\backslash B \right)\cap \left( A\backslash C \right)$.  Hint: First form the universal set consisting of elements of set A, B and C. Find the complement of set B and C i.e. find ${{B}^{C}}$and ${{C}^{C}}$. Take LHS and RHS separately and prove that the simplified set is the same.

First let us form the universal set, which is a set containing all elements of all the sets A, B and C. The universal set is marked by u.

Given set, $A=\left\{ a,b,c,d,e,f,g,x,y,z \right\}$
$B=\left\{ 1,2,c,d,e \right\}$
$C=\left\{ d,e,f,g,2,y \right\}$

We can form the universal set,
$u=\left\{ a,b,c,d,ef,g,x,y,z,1,2 \right\}$

We need to find the complement of B and C. It is represented as ${{B}^{C}}$and ${{C}^{C}}$.

The complement of set B is the set of all elements in the given universal set u that does not belong to set B.

$\therefore {{B}^{C}}=\dfrac{u}{B}$

${{B}^{C}}$contains the elements that do not belong in B but do in the universal set.

$\therefore {{B}^{C}}\left\{ a,b,f,g,x,y,z \right\}$

Similarly, ${{C}^{C}}=\left\{ a,b,c,x,z,1 \right\}$.

Let us first take the LHS of what we have to verify.

$A\backslash \left( B\cup C \right)=A\cap {{\left( B\cup C \right)}^{C}}=A\cap \left( {{B}^{C}}\cap {{C}^{C}} \right)$

{This is the general form of how it is represented}.

Now let’s open the bracket. Doing so, we get,
$=\left( A\cap {{B}^{C}} \right)\cap \left( A\cap {{C}^{C}} \right)$

Now we need to find $\left( A\cap {{B}^{C}} \right)=\left\{ a,b,c,d,e,f,g,x,y,z \right\}\cap \left\{ a,b,f,g,x,y,z \right\}$.

$\left( A\cap {{B}^{C}} \right)$means we need to find the common elements in both set A and set ${{B}^{C}}$.

$\therefore A\cap {{B}^{C}}=\left\{ a,b,f,g,x,y,z \right\}$

Similarly, $A\cap {{C}^{C}}=\left\{ a,b,c,d,e,f,g,x,y,z \right\}\cap \left\{ a,b,c,x,z,1 \right\}$
$=\left\{ a,b,c,x,z \right\}$

$\therefore \left( A\cap {{B}^{C}} \right)\cap \left( A\cap {{C}^{C}} \right)=\left\{ a,b,f,g,x,y,z \right\}\cap \left\{ a,b,c,x,z \right\}$

$=\left\{ a,b,x,z \right\}$

$\therefore$Value of $A\backslash \left( B\cup C \right)=\left\{ a,b,x,z \right\}-(1)$

Now let us consider the RHS$=\left( A\backslash B \right)\cap \left( A\backslash C \right)$

$=\left( A\cap {{B}^{C}} \right)\cap \left( A\cap {{C}^{C}} \right)$

\begin{align} & \left( A\cap {{B}^{C}} \right)=\left\{ a,b,f,g,x,y,z \right\} \\ & \left( A\cap {{C}^{C}} \right)=\left\{ a,b,c,x,z \right\} \\ & \therefore \left( A\cap {{B}^{C}} \right)\cap \left( A\cap {{C}^{C}} \right)=\left\{ a,b,f,g,x,y,z \right\}\cap \left\{ a,b,c,x,z \right\}=\left\{ a,b,x,z \right\} \\ \end{align}

$\therefore$Value of $\left( A\backslash B \right)\cap \left( A\backslash C \right)=\left\{ a,b,x,z \right\}-(2)$.

Comparing (1) and (2) we can verify that,

$A\backslash \left( B\cup C \right)=\left( A\backslash B \right)\cap \left( A\backslash C \right)$

which have the same set $\left\{ a,b,x,z \right\}$.

Hence proved.

Note: You can only solve problems like this if you know the common set symbols used in set theory.
Like $\left\{ {} \right\},A\cup B,A\cap B,{{A}^{C}}$ etc which are used in the question to verify the answer. Remember their basic symbols and what they implement.
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Right Angle  Right Triangle Formula  Isosceles Right Triangle  Right Circular Cone  Right Angle Triangle  Right Circular Cylinder  Right Rectangular Prism  Operation on Sets Intersection of Sets and Difference of Two Sets  Right Angled Triangle Constructions  Right Triangle Congruence Theorem  