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Which of the following sets of component(s) is/are sufficient to implement any arbitrary Boolean function?

  1. XOR gates, NOT gates

  2. $2$ to $1$ multiplexers

  3. AND gates, XOR gates

  4. Three-input gates that output $(A.B) + C$ for the inputs $A, B$ and $C$.

in Digital Logic by Veteran
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4 Answers

+38 votes
Best answer
1. XOR and NOT gates can only make XOR and XNOR which are not functionally complete- $a \oplus \bar a = 1, a \oplus a = 0.$

2. 2-1 multiplexer is functionally complete provided we have external 1 and 0 available. For NOT gate, use $x$ as select line and use 0 and 1 as inputs. For AND gate, use $y$ and 0 as inputs and $x$ as select. With {AND, NOT} any other gate can be made.

3. XOR can be used to make a NOT gate ($a \oplus 1 = \bar a$) and {AND, NOT} is functionally complete. Again this requires external 1.

4. We have $AB + C$. Using $C = 0$, we get an AND gate. Using $B = 1$ we get an OR gate. But we cannot derive a NOT gate here.

So, options B and C are true provided external 1 and 0 are available.
by Veteran
selected by

2)say A,B two inputs , x select line  and A=0,B=1

x'A+xB=x ,so how it is functionally complete?
We are given multiplexers and we can use it any way we want- there is no restriction on how the inputs/select lines are. Our aim is to make some universal gate out of it- then that can be used to make any GATE.
ok some of them , not all
if we make some universal gate- we can make all also :)
yes but not for all input , as that example I given previously it will not work
that doesn't matter- we are given multiplexers and we can connect inputs in any way.
We can determine OR using XOR ((A⊕1).(B⊕1))⊕1= (A'.B')' = A+B

So XOR,NOT is also functionally complete.
how u r getting AND for (A⊕1).(B⊕1) between (A⊕1)   and   (B⊕1)
0 mistake!!

So with the help of 1 and 0 B and C are functionally complete ??

But is it allowed to use 0 or 1 to prove completeness?
yeah it is allowed
even universal logic gates nand,nor cant give 1 or 0

@Arjun Sir

in this link : {AND , EX-OR }  not functionaly complete 

plz confirm !

Isn't that the set of functions which can derive all boolean functions with the help of 1/0, called partially functionally complete??

how can  {∧,⊕} functionally complete either, because if all the inputs are false, then the output is always false too and hence it will preserve the false value, which is contradictory to the property of functionally complete sets??

@sushmita, with XOR gate, if one input is set to 1, then XOR gate behaves as NOT gate.

So, with AND and XOR gate we have AND and NOT gate, which is sufficient to realise any boolean function.
And the questions asks "is /are sufficient to implement any boolean function", so using XOR and AND you can implement any boolean function without using any other logic gate

@Arjun sir.
If help of $0 \ or\  1$ is needed to derive any of $(\wedge,\neg) \ or \  (\vee \neg)$- That function is partially functionally complete.
Still, is it considered as functionally complete?
Like in this question, in both B and C options help of 0 or 1 is required.
And in this question too. But they considered them functionally complete. 


NOT,XOR is not Functionally complete


and OR,XOR is functionally complete

because XOR can itself create NOT gate

because XOR can itself create NOT gate

It is possible with the help of 1. That's what I am saying. If input 1 or 0 is not available then? 


yes u r right.

{XOR,AND} not functionally complete

rather {XOR,AND} and the constant TRUE is functionally complete


{XOR , AND } is partially functionally complete set ,as to derive not we are taking the help of 1 ,

But here in question it is asking that Which of the following sets of component(s) is/are sufficient to implement any arbitrary Boolean function?

So can we say that both B) and C) should be right option?
@prince-that is what arjun sir has said in the answer :)

sets of component(s) is/are sufficient to implement any arbitrary Boolean function?

Does this allow us to use external 0 / 1 ??


@Arjun Sir I think in 4 C=0 will get the AND not C=1. Please correct me if I am wrong 

Yes, you are right.

Multiplexers Are Functionally Complete 

+10 votes
Functionally complete operation set is a set of Logic from which any arbitrary Boolean function can be realized .

Exp of Functionally complete set operation are


PC 2 { NAND }

PC 3 ( NOR )

PC 4 { XOR  AND }  

Note - By Heart  all.
by Boss
MUX too



@shekhar chauhan {XOR, AND} are partially functionally complete. You need external 1 to get the NOT gate.

+1 vote
Answer is C because Multiplexer is the universal logic circuit that Comprise All the boolean function
by Junior
–2 votes
answer - C
by Loyal
I think B ,C

can you tell me please why not B????
2:1 MUX cant implement binary boolean functions
can you please explain why the answer is c ?
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