12/31/2010. Overview. 04-Boolean Algebra Part 2 Text: Unit 2. Basic Theorems. Basic Theorems. Basic Theorems. Examples

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1 Overview 04-Boolean lgebra Part 2 Text: Unit 2 Basic Theorems Multiplying and Factoring ECEGR/ISSC 201 Digital Operations and Computations Winter 2011 Dr. Louie 2 Basic Theorems Basic Theorems Basic laws and theorems of Boolean algebra + 0 = + 1 = 1 x 1 = x 0 = 0 Idempotent laws + = x = = = Dr. Louie 3 Dr. Louie 4 Basic Theorems Examples Involution Law ( ) = Laws of complementarity + = 1 x = 0 x 0 = = Dr. Louie 5 Dr. Louie 6 1

2 Commutative, ssociative and Distributive Laws Commutative laws X = X X + = + X ssociative laws XZ = (X)Z = X(Z) X + +Z = (X + ) + Z = X + ( + Z) Commutative, ssociative and Distributive Laws Distributive laws X( + Z) = X + XZ X + Z = (X +)(X +Z) Second law is not valid in ordinary algebra Dr. Louie 7 Dr. Louie 8 Commutative, ssociative and Distributive Laws Proof: X + Z = (X +)(X +Z) (X +)(X +Z) = XX + XZ + X + Z = X + XZ + X + Z = X1+ XZ + X + Z =X(1 + Z + ) + Z =X + Z X + X = X (X + )(X + ) = X X +X = X X(X+) = X (X+ ) = X X + = X + Dr. Louie 9 Dr. Louie 10 nother simplifying theorem (X+)(X +Z) = XZ+X Proof: (X+)(X +Z) = XZ+X Consider X = 1 (1+)(0+Z) = Z Consider X = 0 (0+)(1+Z) = Therefore: (X+)(X +Z) = XZ+X Dr. Louie 11 Dr. Louie 12 2

3 Recall the statement The light is to turn on if button and button B are pushed or if button is pushed and button B is not pushed or if button C is pushed L = B +B +C can be reduced to: The light turns on if button or button C are pushed L = +C Why? L = B +B +C Using X+X = X => L = + C Dr. Louie 13 Dr. Louie 14 Simplification theorems can be proved using truth tables or the basic laws Example: Use a truth table and Boolean algebra to show that X +X = X X X X +X Example Use a truth table to show X +X = X X X X +X X +X = (X+X)(X+) = X(X +) Considering either value for : X +0 = X, X + 1 =1 XX = X X(1) = X => X(X +) = X Dr. Louie 15 Dr. Louie 16 See page 42 for proofs of other simplification theorems Example: Simplify Z = BC + Z = Set =X; =BC => Z = X + X = X = Simplify Z = [ + B C + D + EF][ + B C + (D + EF) ] Set: X = +B C; = D +EF Z = [X + ][X + ] By second distributive law: Z =[X + ][X + ] = X+ = X = +B C Dr. Louie 17 Dr. Louie 18 3

4 There are many ways of writing equivalent expressions using the simplification theorems We are interested in two particular standard forms Sum-of-products (SoP): an expression that is the sum of products of single variables Examples: B + CD E + C E B + CDE + F Not in SoP form: (+B)CD + EF Multiply out to get into SoP form Dr. Louie 19 Dr. Louie 20 Product-of-sums (PoS): an expression in which the sums are sums of single variables Examples (+B)(C+D +E)(+C +E ) B C(D +E) But not: (+B)(C+D)+EF Factor to get into PoS form Example Identify the following as being in SoP, PoS or an unspecified form F = B + B F = (BC)+(BD) F= B+C+D+E F = (+C)(E) Dr. Louie 21 Dr. Louie 22 Example Identify the following as being in SoP, PoS or an unspecified form F = B + B SoP F = (BC)+(BD) SoP F= B+C+D+E SoP F = (+C)(E) PoS To convert an expression to SoP form, multiply it out Multiplying out usually involves one or more of these laws and simplifications: (X+)(X+Z) = X+Z (X+)(X +Z) = XZ + X X(+Z) = X + XZ Dr. Louie 23 Dr. Louie 24 4

5 In general, when multiplying out avoid unnecessary terms by: First using: (X+)(X+Z) = X+Z and/or (X+)(X +Z) = XZ+X then use: X(+Z) = X + XZ F=(+BC)(+D+E) Setting X = ; = BC; Z = D+E and using (X+)(X+Z) = X+Z F = + BC(D+E) F= + BCD + BCE (using X(+Z) = X + XZ) Dr. Louie 25 Dr. Louie 26 The alternative is multiplying out the original expression and cancelling redundant terms F=(+BC)(+D+E) F = + D + E + BC + BCD + BCE F = + (D + E + BC) + BCD + BCE Using X + X = X F = + BCD + BCE To convert an expression to PoS form, factor it Factoring involves the same equations, but working right to left (X+)(X+Z) = X+Z (X+)(X +Z) = XZ + X X(+Z) = X + XZ Dr. Louie 27 Dr. Louie 28 Factor F=C D +C E + G H F=C (D+E )+G H (using X(+Z) = X + XZ) F=(C (D+E ) + G )(C (D+E )+H) (using (X+)(X+Z) = X+Z) F=((G +C )(G +D+E ))((H+C )(H+D+E )) (using (X+)(X+Z) = X+Z) Write the following in Sum of Product form F=(+B+C )(+B+D)(+B+E)(+D +E)( +C) Dr. Louie 29 Dr. Louie 30 5

6 Want in SoP form, so we need to multiply out First use: (X+)(X+Z) = X+Z (+B+C )(+B+D)(+B+E)(+D +E)( +C) (+B+C )(+B+D)(+B+E)(+D +E)( +C) Can we use it anywhere else? (X+)(X+Z) = X+Z (+B+C D)(+B+E)(+D +E)( +C) (+B+C D)(+B+E)(+D +E)( +C) (+B+C D)(+B+E)(+D +E)( +C) (+B+C DE)(+D +E)( +C) Dr. Louie 31 Dr. Louie 32 We could use it again, but it does not help much Instead look to use: (X+)(X +Z) = XZ+X (+B+C DE)(+ D +E)( +C) (+B+C DE)(+ D +E)( +C) (+B+C DE)(C + (D +E)) Not much more we can do here, so multiply out (+B+C DE)(C + (D +E)) C + (D +E) + BC + B (D +E) + C DEC + C DE (D +E) Simplifying further: C BC + B (D +E) C DE (D +E) Dr. Louie 33 Dr. Louie 34 multiplying out again: C + BC + B (D +E) + C DE (D +E) C + BC + B D + B E + C DE D + C DE E simplifying further C + BC + B D + B E C D E re we done? C + BC + B D + B E + C D E It is in SoP form, but we can minimize it further C + BC + B D + B E + C D E C + B D + B E + C D E using X + X = X Dr. Louie 35 Dr. Louie 36 6

7 If we would have multiplied out the original expression, we would have 3 x 3 x 3 x 3 x 2 = 162 terms! (+B+C )(+B+D)(+B+E)(+D +E)( +C) See text, page 60 for an example of factoring SoP: usually several ND gates input into a single OR gate PoS: usually OR gates input into a single ND gate SoP PoS B C F = B + C B C D F = (+B)(C+D) Dr. Louie 37 Dr. Louie 38 Simplifications See page 52 for a table of laws and theorems of Boolean lgebra Dr. Louie 39 7