Digital Circuit Simplification


This is a Boolean expression simplification exercise.There are several Boolean expressions, functions that produce the same results for all possible values in the truth table. In this article, we consider the equipment minimization criteria. Consider the two functions given below.

F1 = AB’C + A’BC + A’B’C
F2 = A’C + B’C

F2 is the simplified form of F1. Because it has fewer gates and a reduced number of literals. So far, we have studied two ways to deal with circuit complexity in terms of minimum equipment. The two ways are given below.


Reduced number of literals
Fewer possible gates

Boolean Expression Simplification
With the help of solving these examples, you will be able to minimize the expression.

Example 1: A(A + B)


= A.A + A.B Distributive Law

= A + A.B Rule 7

= A (1 + B) Distributive Law

= A (1) Rule 2

= A Rule 4

Example 2: A(A’+AB)

= AA’ + AAB Distributive Law

= 0 + A.B Rule 8, Rule 7

= A.B

Example 3: BC + B’C

= C.(B+B’) Distributive Law

= C.(1) Rule 7

= C Rule 4

Example 4: A(A+A’B)

= A.A + A.A’B Distributive Law

= A + 0.B Rule 7 , Rule 8

= A

Example 5: AB’C + A’BC + A’B’C

= AB’C + A’C(B+B’) Distributive Law

= AB’C + A’C(1). Rule 6

= C (AB’ + A’)

= C (A’ + B’). Rule 11

= A’C + B’C

Example 6: BD + B(D+E) + D'(D+F)

= B (D+D+E) + DD’ + D’F Distributive Law

= B (D + E) + 0 + D’F Rule 5 Rule 8

= BD + BE + D’F

Example 7: A’B’C + (A+B+C’)’ + A’B’C’D

= A’B’C + A’B’C” + A’B’C’D

                                 De Morgan's Theorem

= A’B’C + A’B’C + A’B’C’D Rule 9

= A’B’C + A’B’C’D Rule 5

=A’B’ (C + C’D)

= A’B’ (C + D) Rule 11

Example 8: (B + BC)(B + B’C)(B + D)

= (B)(B + C)(B + D) Rule 10 Rule 11

= (B.B + B.C)(B + D)

= (B + B.C)(B + D) Rule 7

= B (B + D) Rule 10

= B.B + B.D

= B + B.D Rule 7

= B Rule 10

Example 9: A’B’C’D + AB(CD)’ + (AB)’CD

= A’B’C’D + AB(C’ + D’) + (A’ + B’)CD De Morgan’s Theorem

= A’B’C’D + ABC’ + ABD’ + A’CD + B’CD

= A’D(B’C’ + C) + ABC’ + ABD’ + B’CD

= A’D(B’ + C) + ABC’ + ABD’ + B’CD

                                              Rule 11

= A’B’D + A’DC + ABC’ + ABD’ + B’CD

Example 10: ABC[AB + C'(BC + AC)]

= ABC [AB + BCC’ + ACC’]

= ABC [AB + 0 + 0 ] Rule 8

= ABC [AB]

= ABC Rule 7

Example 11: (A + B’)(A + C)

= A.A + A.C + A.B’ + B’C

= A + A.C + A.B’ +B’C Rule 7

= A (1 + C) + A.B’ +B’C

= A.1 + A.B’ +B’C Rule 2

= A ( 1 + B’) + B’C

= A.1 + B’C Rule 2

= A + B’C

Example 12: A’B + A’BC’ + A’BCD + A’BC’D’E

= A’ (B + BC’) + A’BCD + A’BC’D’E

= A’.B + A’BCD + A’BC’D’E Rule 10

=>> Let, A’B = X, CD = Y

= X + XY + XY’E

= X + X(Y+ Y’E)

= X + X(Y+ E) Rule 11

= X + XY + XE

= X (1+Y) + XE

= X (1) + XE Rule 2

= X + XE

= X (1 + E)

= X (1) Rule 2

= X

= A’B X = A’B

If you look at the example, this problem can be solved in two steps with the help of Rule 10.

Let’s solve it in another way.

= A’B + A’BCD + A’BC’D’E Rule 10 on first two terms

= A’B + A’BC’D’E Again Rule 10 on first two terms

= A’B Again Rule 10 on the remaining two terms.

Example 13: AB + A’B’C + A

= AB + A + A’B’C

= A + A’B’C Rule 10

= A + B’C Rule 11

Example 14: (A + A’)(AB + ABC’)

= (1) . A(B + BC’) Rule 6

= A . (B) Rule 10

= A.B

Example 15: AB + (A’ + B’)C + AB

= AB + AB + A’C + B’C

= AB + A’C + B’C. Rule 5

Example 16: (A.B + C.D) [(A’ + B’)(C’ + D’)]
Observing both terms, we find out that both expressions complement each other. Here is an article on how to find the complement of a Boolean expression.

= [(A + B) (C + D)] [(A’ + B’) (C’ + D’)]

=> Let X = (A + B) (C + D)

= X . X’

= 0 Rule 8

Example 17: A.B.C.D + A.B.C’.D’ + A.B.C’.D + A.B.C.D.E + A.B.C’.D’.E’ + A.B.C’.D.E

= A.B ( C.D + C’.D’) + A.B.C’.D + A.B.C.D.E + A.B.C’.D’.E’ + A.B.C’.D.E

= A.B (1) + A.B.C’.D + A.B.C.D.E + A.B.C’.D’.E’ + A.B.C’.D.E Rule 6

= A.B + A.B.C.D.E + A.B.C’.D’.E’ + A.B.C’.D + A.B.C’.D.E

= A.B + A.B (C.D.E + C’.D’.E’) + A.B.C’.D (E + 1). Distributive Law

= A.B + A.B (1) + A.B.C’.D (1) Rule 6 Rule 2

= A.B + A.B + A.B.C’.D

= A.B + A.B.C’.D Rule 5

= A.B Rule 10

Example 18: X’Y’ + XY + X’Y

= X’ (Y’ + Y) + XY

= X’ (1) + XY Rule 6

= X’ + Y Rule 11

Example 19: (X + Y)(X + Y’)

= X.X + X.Y’ + X.Y + Y.Y’ Distributive Law

= X + X(Y’ + Y) + 0 Rule 7 Rule 8

= X + X (1) Rule 6

= X Rule 5

Example 20: X’Y +XY’ + XY + X’Y’

= X’ (Y + Y’) + X (Y + Y’)

= X’ (1) + X (1) Rule 6

= X’ + X

= 1 Rule 6

Example 21: X’ + XY + XZ’ + XY’Z’

= X’ + XY + XZ’ (1 + Y’)

= X’ + Y + XZ’ (1) Rule 11 Rule 2

= X’ + XZ’ + Y

= X’ + Z’ + Y Rule 11

Example 22: XY’ + Y’Z’ + X’Z’

This is the consensus theorem. The answer is XY’ + X’Z’

= XY’ + Y’Z’ (1) + X’Z’ Rule 4

= XY’ + Y’Z’ (X + X’) + X’Z’ Rule 6

= XY’ + Y’ZX + Y’Z’X’ + X’Z’

= XY'(1 + Z) + X’Z’ (1 + Y’). Rule 2

= XY’ + X’Z’

Example 23: ABC + A’B + ABC’

= AB ( C + C’) + A’B

= AB (1) + A’B Rule 6

= AB + A’B

= B (A + A’)

= B (1) Rule 6

= B

Example 24: X’YZ + XZ

= Z (X’Y + X)

= Z (X + Y) Rule 11

= X’Z’ + YZ

Example 25: (X + Y)’ (X’ + Y’)

= (X + Y)'(X’ + Y’)

=X’ Y’ (X’ + Y’)

De Morgan’s Theorem

= X’X’Y’ + X’Y’Y’

= X’Y’ + X’Y’ Rule 7

= X’Y’ Rule 5

Example 26: XY + X(WZ + WZ’)

= XY + XW(Z + Z’)

= XY + XW(1) Rule 6

Example 27: (BC’ + A’D)(AB’ + CD’)

= BC’AB’ + BC’CD’ + A’DAB’ + A’DCD’

= 0 + 0 + 0 + 0 Rule 8

Example 28: A’C’ + ABC + AC’

= A’C’ + AC’ + ABC

= C’ (A’ + A) + ABC

= C’ (1) + ABC. Rule 6

= C’ + AB Rule 11

Example 29: (X’Y’ + Z)’ + Z + XY +WZ

= (X’Y’)’.Z’ + Z + XY + WZ

        De Morgan's Theorem

= (X” + Y”)Z’ + Z + XY + WZ

        De Morgan's Theorem

= (X + Y)Z’ + Z + XY + WZ

= XZ’ + YZ’ + Z + XY + WZ

= XZ’ + YZ’ + XY + Z(W + 1) Rule 2

= XZ’ + YZ’ + XY + Z(1)

= XZ’ + XY + YZ’ + Z Rearrange

= XZ’ + XY + Z + Y Rule 11

= XZ’ + Z + XY + Y Rearrange

= Z + X + Y Rule 11 Rule 10

Example 30: A’B(D’ + C’D) + B (A + A’CD)

= A’B (D’ + C’) + B (A + CD) Rule 11 Rule 11

= A’BD’ + A’BC’ + AB + BCD

= B(A’D’ + A) + A’BC’ + BCD

= B(A + D’) + A’BC’ + BCD Rule 11

= AB + BD’ + A’BC’ + BCD

= B(A + A’C’) + B(D’ + DC)

= B(A + C’) + B(D’ + C) Rule 11 Rule 11

= AB + BC’ +BD’ + BC

= AB + BD’ + B(C’ + C)

= AB + BD’ + B(1) Rule 6

= AB + B (D’ + 1)

= AB + B (1) Rule 2

= B Rule 10

Example 31: (A’ + C)(A’ + C’)(A + B + C’D)

= (A’A’ + A’C’ + A’C + CC’)(A + B + C’D)

= (A’ + A’C’ + A’C + 0)(A + B + C’D)

Rule 7 Rule 8

=(A’ + A’C)(A + B + C’D) Rule 10

= A'(A + B + C’D) Rule 10

= A’A + A’B + A’C’D

= 0 + A’B + A’C’D Rule 8

Example 31: (A’ + C)(A’ + C’)(A + B + C’D)

x=(AB’C + AB’BD + A’B’)C

= (AB’C + 0 + A’B’)C Rule 8

= AB’CC + A’B’C

= AB’C + A’B’C Rule 7

= B’C (A +A’)

= B’C(1) Rule 6

Example 33: [AB (C + (BD)’) + (AB)’] CD

= [AB ( C + B’ + D’) + A’ + B’] CD

De Morgan’s Theorem De Morgan’s Theorem

= [ABC + ABB’ + ABD’ + A’ + B’]CD

= [ABC + A’ + ABD’ + B’ + 0] CD

Rule 8

= [A’ + BC + B’ + AD’] CD Rule 11 Rule 11

= [A’ + AD’ + B’ + BC] CD Rearrange

= [A’ + D’ + B’ + C] CD Rule 11Rule 11

= A’CD + CD’D + B’CD + CCD

= A’CD + 0 + B’CD + CD Rule 8 Rule 7

= A’CD + CD Rule 10

= CD Rule 10

Example 34: A’BC + AB’C’ + A’B’C’ + AB’C + ABC

= BC(A’ + A) + B’C'(A + A’) + AB’C

= BC(1) + B’C'(1) + AB’C Rule 6

= BC + B'(C’ + AC)

= BC + B'(C’ + A) Rule 11

= BC + AB’ + B’C’

Example 35: ABC’ + A’B’C + A’BC + A’B’C’

= ABC’ + A’C(B + B’) + A’B’C’

= ABC’ + A’C(1) + A’B’C’ Rule 6

= ABC’ + A'(C + B’C’)

= ABC’ + A'(C + B’) Rule 11

= ABC’ + A’B’ + A’C

Example 36: (AB + AC)’ + A’B’C

= (A (B + C))’ + A’B’C

= A’ + (B + C)’ + A’B’C

De Morgan’s Theorem

= A’ + B’C’ + A’B’C

De Morgan’s Theorem

= A’ + B'(C’ + A’C)

= A’ + B’ (C’ + A’) Rule 11

= A’ + A’ + B’C’

= A’ + B’C’ Rule 5

Example 37: (AB)’ + (AC)’ + A’ B’C’

= (A’ + B’) + (A’ + C’) + A’B’C’

De Morgan’s Theorem

= A’ + A’ + B’ + C’ + A’B’C’

= A’ + B’ + C’ + A’B’C’ Rule 5

= A’ + B’ + C’ +(A’B’)C’

= A’ + B’ + C’ Rule 10

Example 38: A + AB + AB’C

= A + A(B + B’C)

= A + A(B + C) Rule 11

= A + AB + AC

= A + AC Rule 10

= A Rule 10

Example 39: (A’ + B)C + ABC

= A’C + BC + ABC

= BC + C(A’ + AB)

= BC + C(A’ + B) Rule 11

= BC + A’C + BC

= BC + A’C Rule 5

Example 40: AB’C(BD + CDE) + AC’

= AB’CBD + AB’CCDE + AC’

= 0 + AB’CDE + AC’ Rule 8 Rule 7

= AB’CDE + AC’

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