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 $L.H.S=R.H.S$