Boolean function:

A Boolean function is a function which has one or more binary input and produces a Single binary output. It contain only two values:

• True (1)
• False (0)

Usage in Computer:

There are many Uses of Boolean functions in the Computer for various Operations.

1. Arithmetic Operations:
Boolean functions are used in Arithmetic logic units (ALUs) of CPUs to perform operation like addition, Substraction, multiplication and even division.

2. Data Processing:
Boolean functions are used to process binary data in memory and Storage devices, ensuring efficient data manipulation and retrieval.

3. Control Logic:
Boolean functions are applied in computers to control various parts of the system's Operation to function in co-ordinated manner.

Construction of Truth Table:

The Construction of Boolean expression is done by employing the basic logical operations Such as AND, OR, NOT, which connect the input to generate the Correct output.

Example:

Consider a Boolean function with two inputs A and B. We can construct function F that represents the AND operation:

Function: Y = F(A, B) = A · B

The diagram shown above demonstrates a basic digital Circuit, which is an AND gate.

• The box symbolizes the AND function Y = F(A,B) = A·B.
• This box has two Inputs: A and B.
• Rule: If both A and B are 1, the output will be 1. In any other case, the Output will be 0.
• The Inputs are shown at the entrance of the box, while the output is displayed at the exit of the block.

Truth Table for AND Gate:

Boolean Algebra:

Boolean algebra is the branch of mathematics related to logic and symbolic computation, using two values namely True and False.

Binary variables and Logic operations:

• Binary variables that can have only two values: 0 and 1.
• Logic operations are basic operations implemented in Boolean algebra for processing of these binary Variables.
• The primary logic operations are AND, OR and NOT.

AND Operation:

• AND is the basic logic operator which is used in Boolean algebra.
• It requires two binary inputs which will give a Single binary output.
• The Symbol · is used for an AND operation.
• The output of AND operation is 1 only when both Inputs are 1. Otherwise the result is 0.

Example:

Consider two binary variables:

• A = 1 (True)
• B = 0 (False)

The AND operation can be written as:
          P = A · B

In this example: A = 1 , B = 0. Then, the result P of the AND Operation is 0 (False)

Truth Table for AND Gate:

To simplify the Boolean expression $$A.\bar{B}+\bar{A}.B+A.B$$  using a Karnaugh map:
1. Expression:  $$A.\bar{B}+\bar{A}.B+A.B$$ 
Step 1: Draw K-map for two variables A and B

Step 2: Fill the K-map
Determine the output for each combination of A and B based on the expression:
For A=0 and B=0:
$$F=A.\bar{B}+\bar{A}.B+A.B$$  
$$=0\cdot1+1\cdot0+0\cdot0=0$$ 

For A=0 and B=1:
$$F=A.\bar{B}+\bar{A}.B+A.B$$  
$$=0\cdot0+1\cdot1+0\cdot1=1$$ 

For A=1 and B=0:
$$F=A.\bar{B}+\bar{A}.B+A.B$$ 
$$=1\cdot1+0\cdot0+1\cdot0=1$$ 

For A=1 and B=1:
$$F=A.\bar{B}+\bar{A}.B+A.B$$ 
$$=1\cdot0+0\cdot1+1\cdot1=1$$ 

Step 3: Group the 1s in the K-map
Group adjacent 1s to simplify the expression.

From K-map, we can form groups:

1. Group of two 1s in the second column:
$$\bar{A}.B+A.B$$ 
$$=(\bar{A}+A)\cdot B$$ 
$$=1\cdot B$$ 
$$=B$$ 

2. Group of two 1s in the second row:
$$A.\bar{B}+A.B$$ 
$$=(\bar{B}+B)\cdot A$$ 
$$=1\cdot A$$ 
$$=A$$ 

Final Simplified Expression:
$$F(A,B)=B+A$$ 

Designing a 4-Bit Binary Adder:

A 4-bit binary adder adds two 4-bit binary numbers, A(A3, A2, A1, A0) and B(B3, B2, B1, B0), producing a 4-bit Sum (S3, S2, S1, S0) and a Carry-out (Cout). It combines half-adders (HA) and full adders (FA) for efficient Computation.

Steps in design:

Step 1: Understanding Half-Adders and Full Adders

1. Half-Adder (HA):

• Input: Two bits (A and B)
• Output: Sum(S) and Carry (C)
• Boolean Expression: Sum = A ⊕ B ,  Carry = A . B

Truth Table:

2. Full Adder (FA):

• Inputs: Two bits (A and B) plus a carry-in (Cin)
• Outputs: Sum (S) and Carry-out (Cout)
• Boolean Expressions:  S = A ⊕ B ⊕ Cin     ,   Cout = (A . B) + (Cin . (A ⊕ B))

Truth Table:

Step 2: 4-Bit Adder Construction

1. First Stage (Least Significant Bit):

Use a half-adder (HA) to add the first bits of A and B (A0 and B0)

• Sum S0 = A0 ⊕ B0
• Carry C0 = A0 . B0

2. Remaining Stages (A1, A2, A3):

Use full-adders (FA) to add Subsequent bits along with the Carry-in from the previous stage.

• Sn = An ⊕ Bn ⊕ Cin
• Cout = (An . Bn) + (Cin . (An ⊕ Bn))

Step 3: Final Carry-out:

The Carry-out C3 from the final full-adder is the overall Carry of operation.

Truth Table:

Step 4: Logic Diagram

Stage 1: Connect a half-adder for A0 and B0.
Stage 2: Use full-adder for A1, B1 and Carry from S1
Stages 3 & 4: Repeat for A2, B2 and A3, B3.  

By Combining the outputs of each Stage, the 4-bit adder produces a Sum S3, S2, S1 and the final Carry-out.

F(A,B) = A.B + A. B̄

F(A,B) = A . (B + B̄)        (Distributive Law)

F(A,B) = A . 1                 (Complement Law)

F(A,B) = A                      (Identity Law)  

So,  F(A,B) = A

Use De Morgan's Law to Simplify:

 $$F(A,B,C) = \overline{A+B} + \overline{A.C}$$              

             = Ā. B̄ + Ā + C̄        (by De Morgan's Law)
             = (Ā. B̄) + Ā + C̄
             = Ā + C̄                   (by using absorption law)    

1. $$\overline{A+B}$$  • (A+B)
   = A̅ • B̅ • (A+B)                            (by using De Morgan's law)
   = A̅ • B̅ • A + A̅ • B̅ • B                 (by using distributive law)
   = A̅ • A • B̅ + A̅ • B̅ • B                 (by using commutative law)
   = 0 • B̅ + A̅ • 0                             (by using complement law)
   = 0 + 0                                        (by using null law)
   = 0

2. (A + B̅) • (A̅ + B)

   let.     A + B̅ = C
   = C • (A̅ + B)

   = C • A̅ + C • B                              (by using Distributive Law)
   = (A + B̅) • A̅ + (A + B̅) • B            (putting value of C)

   = (A • A̅) + B̅ • A̅ + A • B + B̅ • B   (by using Distributive Law)
   = 0 + B̅ • A̅ + A • B + 0                 (by using Complement Law: A.A̅ = 0)
   = B̅ • A̅ + A • B                              (by using Identity Law: 0 + X = X)
   = A̅ • B̅ + A • B                              (by using Commutative Law: B̅.A̅ = A̅.B̅)

3.  A + A̅ • (B̅ + C)

   = A + A̅ • B̅ + A̅ • C                       (by using Distributive Law)

   = A + A̅ • (B̅ + C)                           (by using Distributive Law)
   = A + B̅ + C                                   (by using Redundancy Law: A + A̅.X = A + X)

4. $$\overline{A.B}$$  + A.B

           =  $$\overline{A.B}$$ + A·B
           = 1                                                  (by Complement Law: X + X̅ = 1)

        e. (A·B) + (A̅·B̅)

        = (A·B + A̅)·(A·B + B̅)                         (by Distributive Law)
        = (A + A̅)·(B + A̅)·(A + B̅)·(B + B̅)       (by Distributive Law)
        = (1)·(B + A̅)·(A + B̅)·(1)                     (by Complement: X+X̅=1)
        = (A̅ + B)·(A + B̅)                               (Final simplified form)