I think you are basically there, but your table doesn't make much sense. ABCD represents a two's complement value. It IS a two's complement value. Your leftmost column is the bit pattern for -ABCD, which iisn't at issue here. This is the table I was trying to get you to come up with:

V|A|B|C|D||l0|eq0
0|0|0|0|0||0|1
1|0|0|0|1||0|0
2|0|0|1|0||0|0
3|0|0|1|1||0|0
4|0|1|0|0||0|0
5|0|1|0|1||0|0
6|0|1|1|0||0|0
7|0|1|1|1||0|0
-8|1|0|0|0||1|0
-7|1|0|0|1||1|0
-6|1|0|1|0||1|0
-5|1|0|1|1||1|0
-4|1|1|0|0||1|0
-3|1|1|0|1||1|0
-2|1|1|1|0||1|0
-1|1|1|1|1||1|0

 

Ohh you wanted the values, I thought you meant the 2's complement of ABCD. The answer was pretty much infront of me the whole entire tim, thanks WBanh and thatoneguy

 

Just a quick question, here is a table I came up with from a boolean function, where A and B are the sign bits of operand, C is the sign bit of the result and V is overflow. I'm suppose to make this into a 4-1 Mux and just needed some help to see if I'm headed in the right direction.

A | B | C | V
0 | 0 | 0 | 0
0 | 0 | 1 | 1
0 | 1 | 0 | 0
0 | 1 | 1 | 0
1 | 0 | 0 | 0
1 | 0 | 1 | 0
1 | 1 | 0 | 1
1 | 1 | 1 | 0

So, if I spilt this up into 4 parts where S1 = A and S0 = B for my signal lines then for AB = 00, V = C and for AB = 11, V = C'. Then

D3 | C
D2 | 0
D1 | 0
D0 | C'

 

You got it.

Note that, in general, you can use any two of the three inputs for your select signals and then use the third to determine the data input. So you have some options and frequently one of them will result in a "better" circuit ("better" depending on the metrics for "better").

But, in this case, the symmetry is such that all of the options are equivalent and the one you've chosen is the most easily understood for a human.

Good job.