MadeEasy WorkBook: Digital Logic - Multiplexer

Question

Why In' is taken as A'B'C'D'? 

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\[
I_{13} \rightarrow \begin{array}{cccc}
A & B & C & D \\
1 & 1 & 0 & 1
\end{array}
\]
$A^{\prime}=A \oplus B=1 \oplus 1=0$
$B^{\prime}=B \cdot D=1 \cdot 1=1$
$C^{\prime}=\overline{\bar{D}}=D=1$
$D^{\prime}=\overline{C A}=\overline{0.1}=1$
$I_{n} \rightarrow A^{\prime} B^{\prime} C^{\prime} D^{\prime}$
$\begin{array}{llll}0 & 1 & 1 & 1\end{array}$
$\left(A^{\prime} B^{\prime} C^{\prime} D^{\prime}\right)=7 \Rightarrow n=7$

\[
\left.F=(2 C+Z C)=C v_{0}+t_{0}\right)=\bar{C}
\]

Example-4.24
Consider the logic circuit given below:

Input at line $I_{13}$ in $16 \times 1 \mathrm{MUX}$ corresponds to output at line $r_{n}^{\prime}$ of $1 \times 16 \mathrm{De}$ MUX. The value of ' $n$ ' is
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Comments

A',B',C' and D' are not to be confused with complement of A,B,C and D resp.

Instead they should have written S3,S2,S1,S0.

Since $I_{13}$ comes as the o/p of MUX, so we know that select lines of MUX must have been 1101 i.e. A=1,B=1,C=0,D=1.

Now its given that $I_n$ is the corresponding o/p of DeMux.

Lets check the select lines of DeMux:

S3 = A xor B = 1 xor 1 =0

S2=BD=1.1=1

S1=(D')'=D = 1

S0= C nand A = 0 nand 1 = 1

So select lines of DeMux (S3,S2,S1,S0) is 0111 which selects $I_n$ so it must be $(0111)_2=(7)_{10}$

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OK... Thanks