11. What are Demorgan's theorom? Prove algebraically the Demorgan's la

1.	11. What are Demorgan's theorom? Prove algebraically the Demorgan's law.:Answer
Answer» According to De Morgan's first law, the complement of the union of two sets A and B is equal to the intersection of the complement of the sets A and B. proof: DeMorgan's 1st law states X+Y¯¯¯¯¯¯¯¯¯¯¯¯¯¯=X¯¯¯¯⋅Y¯¯¯¯ It is sufficient to prove that (X+Y)+X¯¯¯¯⋅Y¯¯¯¯=1 LHS=Y+(X+X¯¯¯¯⋅Y¯¯¯¯)=Y+X+Y¯¯¯¯=(Y+Y¯¯¯¯)+X=1+X=1=RHS Second Law:: DeMorgan's 2nd Law states that X⋅Y¯¯¯¯¯¯¯¯¯¯¯¯=X¯¯¯¯+Y¯¯¯¯ It is sufficient to prove that X⋅Y+(X¯¯¯¯+Y¯¯¯¯)=1 LHS=Y¯¯¯¯+(X¯¯¯¯+X¯¯¯¯¯¯¯¯¯⋅Y)=Y¯¯¯¯+(X¯¯¯¯+Y)=(Y+Y¯¯¯¯)+X¯¯¯¯=1+X¯¯¯¯=1=RHS

11. What are Demorgan's theorom? Prove algebraically the Demorgan's law.:Answer

Answer»

According to De Morgan's first law, the complement of the union of two sets A and B is equal to the intersection of the complement of the sets A and B.

proof:

DeMorgan's 1st law states X+Y¯¯¯¯¯¯¯¯¯¯¯¯¯¯=X¯¯¯¯⋅Y¯¯¯¯

It is sufficient to prove that (X+Y)+X¯¯¯¯⋅Y¯¯¯¯=1

LHS=Y+(X+X¯¯¯¯⋅Y¯¯¯¯)=Y+X+Y¯¯¯¯=(Y+Y¯¯¯¯)+X=1+X=1=RHS

Second Law:: DeMorgan's 2nd Law states that X⋅Y¯¯¯¯¯¯¯¯¯¯¯¯=X¯¯¯¯+Y¯¯¯¯

It is sufficient to prove that X⋅Y+(X¯¯¯¯+Y¯¯¯¯)=1

LHS=Y¯¯¯¯+(X¯¯¯¯+X¯¯¯¯¯¯¯¯¯⋅Y)=Y¯¯¯¯+(X¯¯¯¯+Y)=(Y+Y¯¯¯¯)+X¯¯¯¯=1+X¯¯¯¯=1=RHS

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