Question

prove and state demorgan law

Answer 1

Proof of De Morgan's Law. ... Definition of De Morgan's law: The complement of the union of two sets is equal to the intersection of their complements and the complement of the intersection of two sets is equal to the union of their complements. These are called De Morgan's laws.

Formal Proof of DeMorgan's Theorems

DeMorgan's Theorems: 
a. (A + B) = A* B 
b. A*B = A + B 
Note: * = AND operation

Proof of DeMorgan's Theorem (b): 
For any theorem X=Y, if we can show that X Y = 0, and that X + Y = 1, then 
by the complement postulates, A A = 0 and A + A = 1, 
X = Y. By the uniqueness of the complement, X = Y. 
Thus the proof consists of showing that (A*B)*( A + B) = 0; and also that (A*B) + ( A + B) = 1.

Prove:(A*B)*( A + B)=0(A*B)*( A + B)=(A*B)*A + (A*B)*B)by distributive postulate=(A*A)*B + A*(B*B)by associativity postulate=0*B + A*0by complement postulate=0 + 0by nullity theorem=0by identity theorem(A*B)*( A + B)=0Q.E.D.

Prove:(A*B) + ( A + B)=1(A*B) + ( A + B)=(A + A + B))*(B + A + B)by distributivity B*C + A = (B + A)*(C + A)(A*B) + ( A + B)=(A + A + B))*(B + B + A)by associativity postulate=(1 + B)*(1 + A)by complement postulate=1*1by nullity theorem=1by identity theorem(A*B) + ( A + B)=1Q.E.D.

Since (A*B)*( A + B) = 0, and (A*B) + ( A+ B) =1, 
A*B is the complement of A + B, meaning that A*B=(A + B)'; 
(note that ' = complement or NOT - double bars don't show in HTML) 
Thus A*B= (A + B)''. 
The involution theorem states that A'' = A. Thus by the involution theorem, (A +B)'' = A + B. 
This proves DeMorgan's Theorem (b). 
DeMorgan's Theorem (a) may be proven using a similar approach.

Answer 2

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