Question

White down the first and second
De Morgan's Theorem​

Answer 1

Explanation:

DeMorgan’s First theorem proves that when two (or more) input variables are AND’edand negated, they are equivalent to the OR of the complements of the individual variables. Thus the equivalent of the NAND function will be a negative-OR function, proving that A.B = A+B. We can show this operation using the following table.

Verifying DeMorgan’s First Theorem using Truth Table

InputsTruth Table Outputs For Each TermBAA.BA.BABA + B0001111010101110011011110000

 

We can also show that A.B = A+B using logic gates as shown.

DeMorgan’s First Law Implementation using Logic Gates



 

The top logic gate arrangement of: A.B can be implemented using a standard NAND gate with inputs A and B. The lower logic gate arrangement first inverts the two inputs producing A and B. These then become the inputs to the OR gate. Therefore the output from the OR gate becomes: A+B

Then we can see here that a standard ORgate function with inverters (NOT gates) on each of its inputs is equivalent to a NANDgate function. So an individual NAND gate can be represented in this way as the equivalency of a NAND gate is a negative-OR.

DeMorgan’s Second Theorem

DeMorgan’s Second theorem proves that when two (or more) input variables are OR’edand negated, they are equivalent to the ANDof the complements of the individual variables. Thus the equivalent of the NORfunction is a negative-AND function proving that A+B = A.B, and again we can show operation this using the following truth table.

Verifying DeMorgan’s Second Theorem using Truth Table

InputsTruth Table Outputs For Each TermBAA+BA+BABA . B0001111011001010101001110000

 

We can also show that A+B = A.B using the following logic gates example.

DeMorgan’s Second Law Implementation using Logic Gates



 

The top logic gate arrangement of: A+B can be implemented using a standard NOR gate function using inputs A and B. The lower logic gate arrangement first inverts the two inputs, thus producing A and B. Thus then become the inputs to the AND gate. Therefore the output from the AND gate becomes: A.B

Then we can see that a standard AND gate function with inverters (NOT gates) on each of its inputs produces an equivalent output condition to a standard NOR gate function, and an individual NOR gate can be represented in this way as the equivalency of a NOR gate is a negative-AND.

Although we have used DeMorgan’s theorems with only two input variables A and B, they are equally valid for use with three, four or more input variable expressions, for example:

For a 3-variable input

 

A.B.C = A+B+C

and also

A+B+C = A.B.C

For a 4-variable input

 

A.B.C.D = A+B+C+D

and also

A+B+C+D = A.B.C.D

and so on.

Answer 2

Explanation:

this answer is correct

hope you understand