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Charles Kime & Thomas Kaminski © 2008 Pearson Education, Inc. (Edited by Dr. Muhamed Mudawar for COE 202 & EE 200 at KFUPM) Boolean Algebra and Logic Gates Logic and Computer Design Fundamentals

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Boolean Algebra and Logic Gates 2 Binary Logic and Gates Binary variables take on one of two values. Logical operators operate on binary values and binary variables. Basic logical operators are the logic functions AND, OR and NOT. Logic gates implement logic functions. Boolean Algebra: a useful mathematical system for specifying and transforming logic functions. We study Boolean algebra as a foundation for designing and analyzing digital systems!

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Boolean Algebra and Logic Gates 3 Binary Variables Recall that the two binary values have different names: True/False On/Off Yes/No 1/0 We use 1 and 0 to denote the two values. Variable identifier examples: A, B, y, z, or X 1 for now RESET, START_IT, or ADD1 later

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Boolean Algebra and Logic Gates 4 Logical Operations The three basic logical operations are: AND OR NOT AND is denoted by a dot (·). OR is denoted by a plus (+). NOT is denoted by an overbar ( ¯ ), a single quote mark (') after, or (~) before the variable.

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Boolean Algebra and Logic Gates 5 Examples: is read “Y is equal to A AND B.” is read “z is equal to x OR y.” is read “X is equal to NOT A.” Notation Examples Note: The statement: 1 + 1 = 2 (read “one plus one equals two”) is not the same as 1 + 1 = 1 (read “1 or 1 equals 1”). BAY yxz AX

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Boolean Algebra and Logic Gates 6 Operator Definitions Operations are defined on the values "0" and "1" for each operator: AND 0 · 0 = 0 0 · 1 = 0 1 · 0 = 0 1 · 1 = 1 OR 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 1 NOT 10 01

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Boolean Algebra and Logic Gates 7 01 10 X NOT XZ Truth Tables Tabular listing of the values of a function for all possible combinations of values on its arguments Example: Truth tables for the basic logic operations: 111 001 010 000 Z = X·Y YX AND OR XYZ = X+Y 000 011 101 111

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Boolean Algebra and Logic Gates 8 Truth Tables – Cont’d Used to evaluate any logic function Consider F(X, Y, Z) = X Y + Y Z XYZX YYY ZF = X Y + Y Z 0000100 0010111 0100000 0110000 1000100 1010111 1101001 1111001

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Boolean Algebra and Logic Gates 9 Using Switches Inputs: logic 1 is switch closed logic 0 is switch open Outputs: logic 1 is light on logic 0 is light off. NOT input: logic 1 is switch open logic 0 is switch closed Logic Function Implementation Switches in series => AND Switches in parallel => OR C Normally-closed switch => NOT

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Boolean Algebra and Logic Gates 10 Example: Logic Using Switches Light is on (L = 1) for L(A, B, C, D) = and off (L = 0), otherwise. Useful model for relay and CMOS gate circuits, the foundation of current digital logic circuits Logic Function Implementation – cont’d B A D C A (B C + D) = A B C + A D

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Boolean Algebra and Logic Gates 11 Logic Gates In the earliest computers, switches were opened and closed by magnetic fields produced by energizing coils in relays. The switches in turn opened and closed the current paths. Later, vacuum tubes that open and close current paths electronically replaced relays. Today, transistors are used as electronic switches that open and close current paths.

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Boolean Algebra and Logic Gates 12 Logic Gate Symbols and Behavior Logic gates have special symbols: And waveform behavior in time as follows : X 0011 Y0101 X · Y(AND)0001 X+ Y(OR)0111 (NOT)X 1100 OR gate X Y Z= X+ Y X Y Z= X · Y AND gate X Z= X NOT gate or inverter

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Boolean Algebra and Logic Gates 13 Logic Diagrams and Expressions Boolean equations, truth tables and logic diagrams describe the same function! Truth tables are unique, but expressions and logic diagrams are not. This gives flexibility in implementing functions. X Y F Z Logic Diagram Logic Equation ZY X F Truth Table 11 1 1 11 1 0 11 0 1 11 0 0 00 1 1 00 1 0 10 0 1 00 0 0 X Y Z Z Y X F

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Boolean Algebra and Logic Gates 14 Gate Delay In actual physical gates, if an input changes that causes the output to change, the output change does not occur instantaneously. The delay between an input change and the output change is the gate delay denoted by t G : tGtG tGtG Input Output Time (ns) 0 0 1 1 0 0.5 1 1.5 t G = 0.3 ns

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Boolean Algebra and Logic Gates 15 1. 3. 5. 7. 9. 11. 13. 15. 17. Commutative Associative Distributive DeMorgan’s 2. 4. 6. 8. X. 1 X = X. 00 = X. XX = 0 = Boolean Algebra 10. 12. 14. 16. X + YY + X = (X + Y)Z + X + (YZ)Z) += X(Y + Z)XYXZ += X + YX. Y = XYYX = (XY)ZX(YX(YZ)Z) = X+ YZ(X + Y)(X + Z)= X. YX + Y = X + 0 X = + X 11 = X + XX = 1 = X = X Invented by George Boole in 1854 An algebraic structure defined by a set B = {0, 1}, together with two binary operators (+ and ·) and a unary operator ( ) Idempotence Complement Involution Identity element

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Boolean Algebra and Logic Gates 16 Some Properties of Boolean Algebra Boolean Algebra is defined in general by a set B that can have more than two values A two-valued Boolean algebra is also know as Switching Algebra. The Boolean set B is restricted to 0 and 1. Switching circuits can be represented by this algebra. The dual of an algebraic expression is obtained by interchanging + and · and interchanging 0’s and 1’s. The identities appear in dual pairs. When there is only one identity on a line the identity is self-dual, i. e., the dual expression = the original expression. Sometimes, the dot symbol ‘ ’ (AND operator) is not written when the meaning is clear

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Boolean Algebra and Logic Gates 17 Example: F = (A + C) · B + 0 dual F = (A · C + B) · 1 = A · C + B Example: G = X · Y + (W + Z) dual G = Example: H = A · B + A · C + B · C dual H = Unless it happens to be self-dual, the dual of an expression does not equal the expression itself Are any of these functions self-dual? (A+B)(A+C)(B+C)=(A+BC)(B+C)=AB+AC+BC Dual of a Boolean Expression (X+Y) · (W · Z) = (X+Y) · (W+Z) (A+B) · (A+C) · (B+C) H is self-dual

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Boolean Algebra and Logic Gates 18 Boolean Operator Precedence The order of evaluation is: 1.Parentheses 2.NOT 3.AND 4.OR Consequence: Parentheses appear around OR expressions Example: F = A(B + C)(C + D)

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Boolean Algebra and Logic Gates 19 Boolean Algebraic Proof – Example 1 A + A · B = A (Absorption Theorem) Proof StepsJustification A + A · B = A · 1 + A · B Identity element: A · 1 = A = A · ( 1 + B) Distributive = A · 1 1 + B = 1 = A Identity element Our primary reason for doing proofs is to learn: Careful and efficient use of the identities and theorems of Boolean algebra, and How to choose the appropriate identity or theorem to apply to make forward progress, irrespective of the application.

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Boolean Algebra and Logic Gates 20 AB + AC + BC = AB + AC (Consensus Theorem) Proof StepsJustification = AB + AC + BC = AB + AC + 1 · BCIdentity element = AB + AC + (A + A) · BCComplement = AB + AC + ABC + ABCDistributive = AB + ABC + AC + ACBCommutative = AB · 1 + ABC + AC · 1 + ACBIdentity element = AB (1+C) + AC (1 + B)Distributive = AB. 1 + AC. 11+X = 1 = AB + ACIdentity element Boolean Algebraic Proof – Example 2

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Boolean Algebra and Logic Gates 21 Useful Theorems Minimization X Y + X Y = Y Absorption X + X Y = X Simplification X + X Y = X + Y DeMorgan’s X + Y = X · Y Minimization (dual) (X+Y)(X+Y) = Y Absorption (dual) X · (X + Y) = X Simplification (dual) X · (X + Y) = X · Y DeMorgan’s (dual) X · Y = X + Y

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Boolean Algebra and Logic Gates 22 Truth Table to Verify DeMorgan’s XYX·YX·YX+YXY X · YX·YX·YX+Y 0000111111 0101100011 1001010011 1111000000 X + Y = X · Y X · Y = X + Y Generalized DeMorgan’s Theorem: X 1 + X 2 + … + X n = X 1 · X 2 · … · X n X 1 · X 2 · … · X n = X 1 + X 2 + … + X n

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Boolean Algebra and Logic Gates 23 Complementing Functions Use DeMorgan's Theorem: 1.Interchange AND and OR operators 2.Complement each constant and literal Example: Complement F = F = (x + y + z)(x + y + z) Example: Complement G = (a + bc)d + e G = (a (b + c) + d) e x zyzyx

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Boolean Algebra and Logic Gates 24 Expression Simplification An application of Boolean algebra Simplify to contain the smallest number of literals (variables that may or may not be complemented) = AB + ABCD + A C D + A C D + A B D = AB + AB(CD) + A C (D + D) + A B D = AB + A C + A B D = B(A + AD) +AC = B (A + D) + A C (has only 5 literals) DCBADCADBADCABA

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Boolean Algebra and Logic Gates 25 Next … Canonical Forms Minterms and Maxterms Sum-of-Minterm (SOM) Canonical Form Product-of-Maxterm (POM) Canonical Form Representation of Complements of Functions Conversions between Representations

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Boolean Algebra and Logic Gates 26 Minterms Minterms are AND terms with every variable present in either true or complemented form. Given that each binary variable may appear normal (e.g., x) or complemented (e.g., ), there are 2 n minterms for n variables. Example: Two variables (X and Y) produce 2 x 2 = 4 combinations: (both normal) (X normal, Y complemented) (X complemented, Y normal) (both complemented) Thus there are four minterms of two variables. YX XY YX YX x

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Boolean Algebra and Logic Gates 27 Maxterms Maxterms are OR terms with every variable in true or complemented form. Given that each binary variable may appear normal (e.g., x) or complemented (e.g., x), there are 2 n maxterms for n variables. Example: Two variables (X and Y) produce 2 x 2 = 4 combinations: (both normal) (x normal, y complemented) (x complemented, y normal) (both complemented) YX YX YX YX

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Boolean Algebra and Logic Gates 28 Two variable minterms and maxterms. The minterm m i should evaluate to 1 for each combination of x and y. The maxterm is the complement of the minterm Minterms & Maxterms for 2 variables xyIndexMintermMaxterm 000m 0 = x yM 0 = x + y 011m 1 = x yM 1 = x + y 102m 2 = x yM 2 = x + y 113m 3 = x yM 3 = x + y

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Boolean Algebra and Logic Gates 29 Minterms & Maxterms for 3 variables M 3 = x + y + zm 3 = x y z3110 M 4 = x + y + zm 4 = x y z4001 M 5 = x + y + zm 5 = x y z5101 M 6 = x + y + zm 6 = x y z6011 1 1 0 0 y 1 0 0 0 x 1 0 1 0 z M 7 = x + y + zm 7 = x y z7 M 2 = x + y + zm 2 = x y z2 M 1 = x + y + zm 1 = x y z1 M 0 = x + y + zm 0 = x y z0 MaxtermMintermIndex Maxterm M i is the complement of minterm m i M i = m i and m i = M i

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Boolean Algebra and Logic Gates 30 Purpose of the Index Minterms and Maxterms are designated with an index The index number corresponds to a binary pattern The index for the minterm or maxterm, expressed as a binary number, is used to determine whether the variable is shown in the true or complemented form For Minterms: ‘1’ means the variable is “Not Complemented” and ‘0’ means the variable is “Complemented”. For Maxterms: ‘0’ means the variable is “Not Complemented” and ‘1’ means the variable is “Complemented”.

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Boolean Algebra and Logic Gates 31 Standard Order All variables should be present in a minterm or maxterm and should be listed in the same order (usually alphabetically) Example: For variables a, b, c: Maxterms (a + b + c), (a + b + c) are in standard order However, (b + a + c) is NOT in standard order (a + c) does NOT contain all variables Minterms (a b c) and (a b c) are in standard order However, (b a c) is not in standard order (a c) does not contain all variables

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Boolean Algebra and Logic Gates 32 Sum-Of-Minterm (SOM) Sum-Of-Minterm (SOM) canonical form: Sum of minterms of entries that evaluate to ‘1’ xyzFMinterm 0000 0011m 1 = x y z 0100 0110 1000 1010 1101m 6 = x y z 1111m 7 = x y z F = m 1 + m 6 + m 7 = ∑ (1, 6, 7) = x y z + x y z + x y z Focus on the ‘1’ entries

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Boolean Algebra and Logic Gates 33 + a b c d Sum-Of-Minterm Examples F(a, b, c, d) = ∑(2, 3, 6, 10, 11) F(a, b, c, d) = m 2 + m 3 + m 6 + m 10 + m 11 G(a, b, c, d) = ∑(0, 1, 12, 15) G(a, b, c, d) = m 0 + m 1 + m 12 + m 15 + a b c da b c d+ a b c d a b c d+ a b c d

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Boolean Algebra and Logic Gates 34 Product-Of-Maxterm (POM) Product-Of-Maxterm (POM) canonical form: Product of maxterms of entries that evaluate to ‘0’ xyzFMaxterm 0001 0011 0100M 2 = (x + y + z) 0111 1000M 4 = (x + y + z) 1011 1100M 6 = (x + y + z) 1111 Focus on the ‘0’ entries F = M 2 ·M 4 ·M 6 = ∏ (2, 4, 6) = (x+y+z) (x+y+z) (x+y+z)

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Boolean Algebra and Logic Gates 35 F(a, b, c, d) = ∏(1, 3, 6, 11) F(a, b, c, d) = M 1 · M 3 · M 6 · M 11 G(a, b, c, d) = ∏(0, 4, 12, 15) G(a, b, c, d) = M 0 · M 4 · M 12 · M 15 Product-Of-Maxterm Examples (a+b+c+d)

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Boolean Algebra and Logic Gates 36 Observations We can implement any function by "ORing" the minterms corresponding to the ‘1’ entries in the function table. A minterm evaluates to ‘1’ for its corresponding entry. We can implement any function by "ANDing" the maxterms corresponding to ‘0’ entries in the function table. A maxterm evaluates to ‘0’ for its corresponding entry. The same Boolean function can be expressed in two canonical ways: Sum-of-Minterms (SOM) and Product-of- Maxterms (POM). If a Boolean function has fewer ‘1’ entries then the SOM canonical form will contain fewer literals than POM. However, if it has fewer ‘0’ entries then the POM form will have fewer literals than SOM.

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Boolean Algebra and Logic Gates 37 Converting to Sum-of-Minterms Form A function that is not in the Sum-of-Minterms form can be converted to that form by means of a truth table Consider F = y + x z xyzFMinterm 0001m 0 = x y z 0011m 1 = x y z 0101m 2 = x y z 0110 1001m 4 = x y z 1011m 5 = x y z 1100 1110 F = ∑(0, 1, 2, 4, 5) = m 0 + m 1 + m 2 + m 4 + m 5 = x y z + x y z + x y z + x y z + x y z

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Boolean Algebra and Logic Gates 38 Converting to Product-of-Maxterms Form A function that is not in the Product-of-Minterms form can be converted to that form by means of a truth table Consider again: F = y + x z xyzFMinterm 0001 0011 0101 0110M 3 = (x+y+z) 1001 1011 1100M 6 = (x+y+z) 1110M 7 = (x+y+z) F = ∏(3, 6, 7) = M 3 · M 6 · M 7 = (x+y+z) (x+y+z) (x+y+z)

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Boolean Algebra and Logic Gates 39 Conversions Between Canonical Forms F = m 1 +m 2 +m 3 +m 5 +m 7 = ∑(1, 2, 3, 5, 7) = x y z + x y z + x y z + x y z + x y z F = M 0 · M 4 · M 6 = ∏(0, 4, 6) = (x+y+z)(x+y+z)(x+y+z) xyzFMintermMaxterm 0000M 0 = (x + y + z) 0011m 1 = x y z 0101m 2 = x y z 0111m 3 = x y z 1000M 4 = (x + y + z) 1011m 5 = x y z 1100M 6 = (x + y + z) 1111m 7 = x y z

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Boolean Algebra and Logic Gates 40 Algebraic Conversion to Sum-of-Minterms Expand all terms first to explicitly list all minterms AND any term missing a variable v with (v + v) Example 1: f = x + x y(2 variables) f = x (y + y) + x y f = x y + x y + x y f = m 3 + m 2 + m 0 = ∑(0, 2, 3) Example 2: g = a + b c(3 variables) g = a (b + b)(c + c) + (a + a) b c g = a b c + a b c + a b c + a b c + a b c + a b c g = a b c + a b c + a b c + a b c + a b c g = m 1 + m 4 + m 5 + m 6 + m 7 = ∑ (1, 4, 5, 6, 7)

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Boolean Algebra and Logic Gates 41 Algebraic Conversion to Product-of-Maxterms Expand all terms first to explicitly list all maxterms OR any term missing a variable v with v · v Example 1: f = x + x y(2 variables) Apply 2 nd distributive law: f = (x + x) (x + y) = 1 · (x + y) = (x + y) = M 1 Example 2: g = a c + b c + a b(3 variables) g = (a c + b c + a) (a c + b c + b)(distributive) g = (c + b c + a) (a c + c + b)(x + x y = x + y) g = (c + b + a) (a + c + b)(x + x y = x + y) g = (a + b + c) (a + b + c) = M 5. M 2 = ∏ (2, 5)

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Boolean Algebra and Logic Gates 42 Function Complements The complement of a function expressed as a sum of minterms is constructed by selecting the minterms missing in the sum-of-minterms canonical form Alternatively, the complement of a function expressed by a Sum of Minterms form is simply the Product of Maxterms with the same indices Example: Given F(x, y, z) = ∑ (1, 3, 5, 7) F(x, y, z) = ∑ (0, 2, 4, 6) F(x, y, z) = ∏ (1, 3, 5, 7)

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Boolean Algebra and Logic Gates 43 Summary of Minterms and Maxterms There are 2 n minterms and maxterms for Boolean functions with n variables. Minterms and maxterms are indexed from 0 to 2 n – 1 Any Boolean function can be expressed as a logical sum of minterms and as a logical product of maxterms The complement of a function contains those minterms not included in the original function The complement of a sum-of-minterms is a product-of- maxterms with the same indices

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Boolean Algebra and Logic Gates 44 Standard Sum-of-Products (SOP) form: equations are written as an OR of AND terms Standard Product-of-Sums (POS) form: equations are written as an AND of OR terms Examples: SOP: POS: These “mixed” forms are neither SOP nor POS Standard Forms B C B A C B A C · ) C B (A · B) (A C) (A C) B (A B) (A C A C B A

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Boolean Algebra and Logic Gates 45 Standard Sum-of-Products (SOP) A sum of minterms form for n variables can be written down directly from a truth table. Implementation of this form is a two-level network of gates such that: The first level consists of n-input AND gates The second level is a single OR gate This form often can be simplified so that the corresponding circuit is simpler.

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Boolean Algebra and Logic Gates 46 A Simplification Example: Writing the minterm expression: F = A B C + A B C + A B C + ABC + ABC Simplifying: F = A B C + A (B C + B C + B C + B C) F = A B C + A (B (C + C) + B (C + C)) F = A B C + A (B + B) F = A B C + A F = B C + A Simplified F contains 3 literals compared to 15 Standard Sum-of-Products (SOP) )7,6,5,4,1()C,B,A(F

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Boolean Algebra and Logic Gates 47 AND/OR Two-Level Implementation The two implementations for F are shown below It is quite apparent which is simpler!

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Boolean Algebra and Logic Gates 48 SOP and POS Observations The previous examples show that: Canonical Forms (Sum-of-minterms, Product-of-Maxterms), or other standard forms (SOP, POS) differ in complexity Boolean algebra can be used to manipulate equations into simpler forms Simpler equations lead to simpler implementations Questions: How can we attain a “simplest” expression? Is there only one minimum cost circuit? The next part will deal with these issues

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Boolean Algebra and Logic Gates 49 Terms of Use All (or portions) of this material © 2008 by Pearson Education, Inc. Permission is given to incorporate this material or adaptations thereof into classroom presentations and handouts to instructors in courses adopting the latest edition of Logic and Computer Design Fundamentals as the course textbook. These materials or adaptations thereof are not to be sold or otherwise offered for consideration. This Terms of Use slide or page is to be included within the original materials or any adaptations thereof.

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