3690884862 - Chapter 2 Boolean Expression Simplification...

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CEG 260, Meilin Liu 1 Chapter 2 Boolean Expression Simplification using BooleanAlgebra Meilin Liu Department of Computer Science Wright State University Homepage: http://www.wright.edu Email: meilin.liu@wright.edu
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CEG 260, Meilin Liu 2 1. 3. 5. 7. 9. 11. 13. 15. 17. Commutative Associative Distributive DeMorgan’s 2. 4. 6. 8. X . 1 X = X . 0 0 = X . X X = 0 = X . X Boolean Algebra 10. 12. 14. 16. X + Y Y + X = ( X + Y ) Z + X + ( Y Z ) + = X ( Y + Z ) XY XZ + = X + Y X . Y = XY YX = ( XY ) Z X ( YZ ) = X + YZ ( X + Y )( X + Z ) = X . Y X + Y = X + 0 X = + X 1 1 = X + X X = 1 = X + X X = X
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CEG 260, Meilin Liu 3 Boolean Expression Simplification by Boolean Algebra (1)X+XY = X Proof: X+XY = X*1 +XY = X(1+Y) = X (2)XY+XY’ = X Proof: XY+XY’ = X(Y+Y’) = X*1 = X (3)X+X’Y = X+Y Proof: X+X’Y = (X+X’)(X+Y) = X+Y (4)X(X+Y) = X Proof: X(X+Y) = X+XY = X*1+X*Y = X (1+Y) =X*1=X Extension: a(a+b+c)=aa+ab+ac=a+ab+ac=a*(1+b+c)=a
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CEG 260, Meilin Liu 4 Boolean Expression Simplification by Boolean Algebra (5) (X+Y)(X+Y’) = X Proof: (X+Y)(X+Y’) = X+(Y*Y’) = X +0=X 2 nd Solution: (X+Y)(X+Y’)=X*X+XY’+Y*X+Y*Y’= X*1+XY’+XY+0=X*(1+Y’+Y)=X*1=X (6) X(X’+Y)=XY Proof: X(X’+Y) = (X*X’) +XY = 0+ XY=XY By taking duals on the both sides of equation (1),(2), (3), we got Equation (4), (5),(6)
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3690884862 - Chapter 2 Boolean Expression Simplification...

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