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Chapter 11: Problem 17

Write \(x y, x \vee y, \bar{x},\) and \(x \uparrow y\) in terms of \(\downarrow\)

Short Answer

Expert verified
AND, OR, NOT and NAND operators can be represented using NOR as follows: \(x \land y = \overline{\bar{x} \downarrow \bar{y}}\), \(x \lor y = \overline{\bar{x} \downarrow \bar{x}} \downarrow \overline{\bar{y} \downarrow \bar{y}}\), \(\neg x = \overline{\bar{x} \downarrow \bar{x}}\), and \(x \uparrow y = \overline{\bar{x} \downarrow \bar{y}}\).

Step by step solution

01

Transforming AND Operator

To transform AND operator, \(x \land y\), into NOR, use De Morgan's laws that state \(\overline{A + B} = \bar{A} \land \bar{B}\). By inverting both sides (for notation swap \(\overline{ }\) to \( \)), we get \(A + B = \overline{\bar{A} \land \bar{B}}\). Replacing \(+\) with \(\downarrow\) and using this formula we get: \(x \land y = \overline{\bar{x} \downarrow \bar{y}}\).
02

Transforming OR Operator

To transform OR operator, \(x \lor y\), into NOR, we just need to notice that \(x \lor y = \overline{\bar{x} \downarrow \bar{x}} \downarrow \overline{\bar{y} \downarrow \bar{y}}\).
03

Transforming NOT Operator

To transform NOT operator, \(\neg x\), into NOR, recall that it can also be expressed as \(x \lor x\), which, using our transformation of OR gives: \(\neg x = \overline{\bar{x} \downarrow \bar{x}}\).
04

Transforming NAND Operator

To transform NAND operator, \(x \uparrow y\), into NOR, we can notice that \(\neg(x \land y)\) is equivalent to NAND, so using our previous transformation of AND gives: \(x \uparrow y = \overline{\bar{x} \downarrow \bar{y}}\) .

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boolean Algebra
Boolean algebra is a branch of algebra centered around variables that take on one of two distinct values: true or false. Though these values can also be represented as 0 or 1, respectively. This form of algebra is fundamental in computer science, particularly in the design and analysis of digital circuits. Boolean algebra allows us to rigorously manipulate logical statements using operators such as AND ( AND ), OR ( OR ), and NOT ( NOT ).
  • AND: This operator (also called conjunction) returns true if both operands are true.
  • OR: This operator (or disjunction) returns true if at least one operand is true.
  • NOT: This operator (or negation) inverts the value of the operand.
Understanding these basic operators can help simplify complex logical expressions. They can be transformed and expressed through different operators, like NOR or NAND, to suit specific needs in digital logic design.
Logic Gates
Logic gates are physical devices that implement Boolean operations in digital circuits. Each gate corresponds to a fundamental Boolean operator and performs a specific logical function on one or more input signals.
  • AND Gate: Gives a high output (1) when all inputs are high.
  • OR Gate: Outputs high if at least one input is high.
  • NOT Gate: Inverts the input signal, turning 1 to 0 or vice versa.
In digital electronics, these gates form the building blocks of circuits for computation, permitting complex processes by combining different gate types. They can be further configured using combinations to create alternative logic types such as NAND and NOR gates, which we will explore later.
De Morgan's Laws
De Morgan's Laws are two fundamental rules about complementary duals in Boolean algebra. These laws simplify expressions involving NOT, AND, and OR operations, making logical statements easier to manage.
  • The first law states: \[\overline{A \land B} = \overline{A} \lor \overline{B}\]Which implies that the negation of an AND operation is equivalent to the OR of the negations.
  • The second law states: \[\overline{A \lor B} = \overline{A} \land \overline{B}\]This means the negation of an OR operation equals the AND of the negations.
These transformations are incredibly useful when manipulating expressions during proof construction or circuit design and are key to converting between AND, OR, NOR, and NAND operations.
NOR and NAND Operators
NOR and NAND are specialized logic gates derived from the basic Boolean operators. What makes them special is their versatility; either one can form the basis for all other logic functions.
  • NOR Gate:
    - This gate functions as the inverse of the OR operation. It outputs true only when all inputs are false. - A NOR gate can be used to create NOT, AND, and OR gates through combinations.
  • NAND Gate:
    - Acting inversely to AND, a NAND gate gives a true output unless all inputs are true. - Like the NOR gate, combinations of NAND gates can yield any other logic gate.
    • NOR and NAND gates provide essential flexibility in logical circuit design. For instance, using these gates alone, one can construct entire systems due to their completeness properties. This is particularly useful in both theoretical explorations and practical applications, such as compact and efficient circuit design.

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    Most popular questions from this chapter

    Find the disjunctive normal form of each func. tion using algebraic techniques. (We abbreviate \(a \wedge b\) as \(a b .)\) \(f(x, y, z)=x \vee y(x \vee \bar{z})\)

    Show that the set of gates \\{ \(\mathrm{NOR}\\}\) is functionally complete.

    Tell whether the given expression is a Boolean expression. If it is a Boolean expression, use Definition 11.1 .9 to show that it is. $$ x_{1} \wedge \bar{x}_{2} \vee x_{3} $$

    Let \(*\) be a binary operator on a set \(S\) containing 0 and \(1 .\) Write a set of axioms for \(*\), modeled after rules that NAND satisfies, so that if we define $$ \begin{aligned} \bar{x} &=x * x \\ x \vee y &=(x * x) *(y * y) \\ x \wedge y &=(x * y) *(x * y) \end{aligned} $$ then \((S, \vee, \wedge,-, 0,1)\) is a Boolean algebra.

    Let \(B(x, y)\) be a Boolean expression in the variables \(x\) and \(y\) that uses only the operator \(\leftrightarrow\) (see Definition 1.3 .8 ). (a) Show that if \(B\) contains an even number of \(x\) 's, the values of \(B(\bar{x}, y)\) and \(B(x, y)\) are the same for all \(x\) and \(y\) (b) Show that if \(B\) contains an odd number of \(x\) 's, the values of \(B(\bar{x}, y)\) and \(\overline{B(x, y)}\) are the same for all \(x\) and \(y\) (c) Use parts (a) and (b) to show that \(\\{\leftrightarrow\\}\) is not functionally complete. This exercise was contributed by Paul Pluznikov.
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