A Boolean function, $f(x,y,z)$ with $x$ as MSB and $z$ as LSB is realized by $4:1$ multiplexer (MUX) with select lines, $S_1$ and $S_0$ ($S_1$ is MSB, $S_0$ is LSB) and inputs, $I_0,I_1,I_2,I_3$ as shown in the Figure. Which of the following options is the correct expression of $f(x,y,z)$ ?

The correct option is xz + y.

Step 1: Understand the 4:1 Multiplexer (MUX) Equation
A 4:1 multiplexer selects one of the four inputs $I_0,I_1,I_2,I_3$ based on the values of the two select lines $S_1$ (MSB) and $S_0$ (LSB). The general Boolean expression for the output function is:
$$f(x,y,z)=\overline{S_1}\overline{S_0}{I}_0+\overline{S_1}{S}_0{I}_1+S_1\overline{S_0}{I}_2+S_1{S}_0{I}_3$$

Step 2: Identify the Select Lines and Inputs from the Circuit Diagram
Looking at the provided image:
1. The select lines are connected as:
$$S_1=x$$
$$S_0=z$$
2. Examining the input terminals:
- The variable input $y$ is connected to the inputs $I_0$, $I_1$, and $I_2$.
- The constant input $1$ is connected to the input $I_3$.
Therefore, we have:
$$I_0=y,I_1=y,I_2=y,I_3=1$$

Step 3: Substitute the Values and Simplify the Boolean Expression
Substitute the select lines and inputs into the multiplexer output equation:
$$f(x,y,z)=\overline{x}\overline{z}\left(y\right)+\overline{x}z\left(y\right)+x\overline{z}\left(y\right)+xz\left(1\right)$$
Group the terms containing $y$:
$$f(x,y,z)=y(\overline{x}\overline{z}+\overline{x}z+x\overline{z})+xz$$
Simplify the expression inside the parentheses:
$$\overline{x}\overline{z}+\overline{x}z+x\overline{z}=\overline{x}(\overline{z}+z)+x\overline{z}$$
Using the Boolean identity $\overline{z}+z=1$:
$$=\overline{x}\left(1\right)+x\overline{z}=\overline{x}+x\overline{z}$$
By the distributive law, $A+\overline{A}B=A+B$:
$$\overline{x}+x\overline{z}=\overline{x}+\overline{z}$$
Substitute this back into the main expression:
$$f(x,y,z)=y(\overline{x}+\overline{z})+xz=\overline{x}y+y\overline{z}+xz$$
To simplify $\overline{x}y+y\overline{z}+xz$, we can factor out $y$:
- If $y=1$, then the expression evaluates to $\overline{x}+\overline{z}+xz=(\overline{x}+xz)+\overline{z}=\overline{x}+z+\overline{z}=1$.
- If $y=0$, then the expression evaluates to $xz$.
Thus, we can write the output logically as:
$$f(x,y,z)=y+\overline{y}xz$$
Applying the rule $A+\overline{A}B=A+B$ once again:
$$(x,y,z)=xz+y$$