boolean logic - Combinational circuit: binary output one higher than input -

the question goes follows:

design combinational circuit 3 inputs x, y, z, , 3 outputs a, b, c. when binary input 0, 1, 2, or 3, binary output 1 greater input. when binary input 4, 5, 6, or 7, binary output 2 less input.

i'll honest in have not slightest clue on how start this. i'm guessing i'll using full adder of sort, being can add more 2 bits. first year of electrical engineering @ college, and, sadly, professor not great of teacher...

any extremely appreciated, thanks!

you not need create adder.

to start, draw table of inputs , output:

| *inputs * || * outputs * | | x | y | z || | b | c   | |=========================| | 0 | 0 | 0 || 0 | 0 | 1  | | 0 | 0 | 1 || 0 | 1 | 0  | | 0 | 1 | 0 || 0 | 1 | 1  | | 0 | 1 | 1 || 1 | 0 | 0  | | 1 | 0 | 0 || 0 | 1 | 0  | | 1 | 0 | 1 || 0 | 1 | 1  | | 1 | 1 | 0 || 1 | 0 | 0  | | 1 | 1 | 1 || 1 | 0 | 1  | 

from here, should able create complicated formula each value of a, b, or c.

each of these complicated formulas should simplified as possible. example, can simplify a down to

y , (z or x) 

once have simplest possible formulas 3 outputs, can turn circuit.