The **proof** of this statement :

The 1's complement of an Excess- 3 number is the Excess- 3 code for the 9's complement of the corresponding decimal number.

Let's $x$ is the number that we are considering here ($0 \leq x \leq 9$ for Excess-3 rest are Pseudo-tetrade)

Assuming $x$ is representated in **decimal**

Then the above statement can be written as :

$1's\ Complement(Excess3(x))=Excess3(9's\ Complement(x)) $

Since $x$ is in **decimal**, $Excess3(x)=x+3$

$9's\ Complement(x)=9-x$

and $1's\ Complement(x)=15-x$ [Since $0 \leq x \leq 9$ so it is representated using 4 bits hence for finding $1's\ Complement$ we need to substract **binary x** from **binary 1111** .**i.e 15 - x ** in **decimal **)

Now we can write the

$L.H.S = 1's\ Complement(Excess3(x))$

$\Rightarrow 1's\ Complement(x+3)$

$\Rightarrow 15-(x+3)$

$\Rightarrow 12-x$

$R.H.S=Excess3(9's\ Complement(x))$

$\Rightarrow Excess3(9-x)$

$\Rightarrow (9-x)+3$

$\Rightarrow 12-x $

**Hence **$L.H.S=R.H.S$