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 32: [AB'(C + BD) + A'B']C
=(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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