A **set of operations** is said to be functionally complete or universal if and only if **every** switching function can be expressed by means of operations in it. A set of Boolean functions is functionally complete, if all other Boolean functions can be constructed from this set and a set of input variables are provided, e.g.

- Set A = {+,*,’ (OR, AND, complement) } are functionally complete.
- Set B = {+,’} are functionally complete
- Set C = {*,’} are functionally complete

**Post’s Functional Completeness Theorem –** Important closed classes of functions:

- T
_{0}– class of all 0-preserving functions, such as f(0, 0, … , 0) = 0. - T
_{1}– class of all 1-preserving functions, such as f(1, 1, … , 1) = 1. - S – class of self-dual functions, such as f(x
_{1}, … ,x_{n}) = ¬ f(¬x_{1}, … , ¬x_{n}). - M – class of monotonic functions, such as : {x
_{1}, … ,x_{n}} ≤ {x_{1}, … ,x_{n}}, if x_{i}≤ y_{i}

if {x_{1}, … ,x_{n}} ≤ {x_{1}, … ,x_{n}}

then f(x_{1}, … ,x_{n}) ≤ f(x_{1}, … ,x_{n}) - L – class of linear functions, which can be presented as: f(x
_{1}, … ,x_{n}) = a_{0}+ a_{1}·x_{1}+ … + a_{n}·x_{n}; a_{i}{0, 1}.

**Theorem –** A system of Boolean functions is functionally complete if and only if for each of the five defined classes T_{0}, T_{1}, S, M, L, there is a member of F which does not belong to that class.

These are minimal functionally complete operator sets –

**One element –**

{↑}, {↓}.

**Two elements –**

**Three elements –**

**Examples on functional Completeness – **

**Check if function F(A,B,C) = A’+BC’ is functionally complete?****Explanation –**Let us start by putting all variables as ‘A’ so it becomes

F(A,A,A) = A’+A.A’ = A’—-(i)

F(B,B,B) = B’+B.B’ = B’—(ii)

Now substitute F(A,A,A) in place of variable ‘A’ and F(B,B,B) in place of variable ‘C’

F(F(A,A,A),B,F(B,B,B)) = (A’)’+B.(B’)’ = A+B—(iii)

from (i) and (ii) complement is derived and from (iii) operator ‘+’ is derived so this function is functionally complete as from above if function contains {+,’} is**functionally complete**.**Check if function F(A,B) = A’+B is functionally complete?****Explanation –**Let us start by putting all variables as ‘A’ so it becomes

F(A,A) = A’+A = 1—-(i)

F(B,B) = B’+B = 1—(ii)

F(A,0) = A’+0 = A’—(iv)

Now substitute F(A,0) in place of variable ‘A’

F(F(A,0),B) = (A’)’+B = A+B—(iii)

from (iv) complement is derived and from (iii) operator ‘+’ is derived so this function is functionally complete as from above if function contains {+,’} is**partially functionally complete**.**Check if function F(A,B) = A’B is functionally complete?****Explanation –**Let us start by putting all variables as ‘A’ so it becomes

F(A,A) = A’.A’ = 0—-(i)

F(A,0) = A’.0 = 0—(ii)

F(A,1) = A’.1 = A’—(iv)

Now substitute F(A,1) in place of variable ‘A’

F(F(A,1),B) = (A’)’*B = A*B—(iii)

from (iv) complement is derived and from (iii) operator ‘*’ is derived so this function is functionally complete as from above if function contains {*,’} is**partially functionally complete**.**Note –**If the function becomes functionally complete by substituting ‘0’ or ‘1’ then it is known as partially functionally complete.**Check if function F(A,B) = A’B+AB’ (EX-OR) is functionally complete?****Explanation –**Let us start by putting all variables as ‘A’ so it becomes

F(A,1) = A’.1 + A.0 = A’—-(i)

F(A’,B) = AB + A’B’–(ii)

F(A’,B’) = AB’ + A’B–(iii)

F(A,B’) = A’B’ + AB—(iv)

So there is no way to get {+,*,’} according to condition. So**EX-OR is non functionally complete**.**Consider the operations**

f(X, Y, Z) = X’YZ + XY’ + Y’Z’ and g(X′, Y, Z) = X′YZ + X′YZ′ + XY

Which one of the following is correct?

(A) Both {f} and {g} are functionally complete

(B) Only {f} is functionally complete

(C) Only {g} is functionally complete

(D) Neither {f} nor {g} is functionally complete**Explanation –**See GATE CS 2015 (Set 1) | Question 65

**References –**

Post’s Functional Completeness Theorem

Functional completeness – Wikipedia

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