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I am trying to solve the following combination,

A | B | Cin | Sum | Cout
0   0    0     0     0            
0   0    1     1     0        
0   1    0     1     0         
0   1    1     0     1          
1   0    0     1     0       
1   0    1     0     1
1   1    0     0     1
1   1    1     1     1

Now the equation of Sum using SOP becomes,

ab'Cin'+ a'b'Cin + abCin + a'bCin'

Its been solved to,

a XOR b XOR Cin 

in the solution. Please guide me how can I end up with the result above? Thanks in advance.

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If you know the solutions and want to prove it, you should start from the XOR form and develop it. If not, you can still notice that Sum is the sum of A, B and Cin mod 2. That's what XOR does A XOR B XOR ... XOR Z = A + B + ... + Z mod 2. –  xavierm02 May 7 '13 at 11:41
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1 Answer

up vote 1 down vote accepted

$$ab'C_{in}'+ a'b'C_{in} + abC_{in} + a'bC_{in}'$$ $$C_{in}'(ab'+a'b) + C_{in}(a'b'+ab)$$ This is of the form: $$X'Y + XY'$$ Which is: $$X \oplus Y$$ where, $$X = C_{in}$$ $$Y = (ab'+a'b) = (a \oplus b)$$

So finally you arrive at: $$ a \oplus b \oplus C_{in}$$

Hope the answer is clear !!

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Thanks. Thats quite helpful. –  Fahad Uddin May 7 '13 at 14:52
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