How to prove demorgan theorem algebricaly?
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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.
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.
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