Sum of Palindromes a Palindrome?

Sorry for weak language abilities ... $\large{\begin{array}{ccccc} && & A &B&B&A\\ +& & & C&D&D&C\\ \hline &&E&F&G&F&E\\ \end {array}}$

There are only two different cases for $E,F,G$ ,Let's prove this ...

First case :

I will start with the left side ..

1) E=1, because if E=0, then the palindrome isn't five-digit palindrome, E cant be greater than 1 , because B,D are one-digit numbers , so there sum will leave a carry of maximum 1 ... add this 1 to A plus C you will again that the carry is maximum 1 .

2)A+C=11 not 1, because if A+C=1 , then if we add the carry of B+D, we will get maximum 2 , and E will be 0 . Of course if A+C>11, they must be 1+10k, because if the ones digit we have E which is equal to 1.. and the sum of two one-digit integers cant reach 20 even with carry of one like i said .

IN THIS CASE( FIRST CASE ) B+D+1<10 so i can say the next ..

3) so in the second coloumn we have to add 1+B+D=F .

4) now we dont have carry because the last sum was <10. , so B+D=G which must be <10 also according to 3 .

5) so the last coloumn will be without carry from the previous sum , so A+C= 10+F , the 10 came from the E in the last digit .. But we know that A+C= 11 , so F=1 .

We got : E=1 , F=1, G=F - 1=0( look at 3 and 4 ) .. so the palindrome is 11011 .

Second case:

Continue after number 2 ,assume that 1+B+D $\geq 10$ , so

3) 1+B+D= 10+F

4) according to 3 we now have carry , also 1+B+D (\geq)10, (according to 3) , then 1+B+D=10+G.

5) also now we have carry , so 1+A+C= EF , but A+C=11 , from number 2 , so F=2 .

We got : E=1 , F=2 , G=F(look at 3 and 4 ) .. so the palindrome is 12221 .

The sum is 11011+12221=23232