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Prove that a combinatorial network for $$ (x \wedge y \wedge z) \vee(\neg x \wedge y \wedge z) \vee(x \wedge \neg y \wedge z) \vee(x \wedge y \wedge \neg z) $$ can be simplified to a combinatorial network representing $$ (x \wedge y) \vee(x \wedge z) \vee(y \wedge z) $$

Short Answer

Expert verified
The expression simplifies to \((x \wedge y) \vee (x \wedge z) \vee (y \wedge z)\) using boolean algebra rules.

Step by step solution

01

Understand the Problem

The expression is a boolean logic expression, which means we need to simplify it to another equivalent expression using boolean algebra rules. The goal is to rewrite the original expression \((x \wedge y \wedge z) \vee (eg x \wedge y \wedge z) \vee (x \wedge eg y \wedge z) \vee (x \wedge y \wedge eg z)\) into the simpler form \((x \wedge y) \vee (x \wedge z) \vee (y \wedge z)\).
02

Analyze the Original Expression

The original expression is composed of four terms that each represent a condition that makes the whole expression true: 1. \(x \wedge y \wedge z\) 2. \(eg x \wedge y \wedge z\) 3. \(x \wedge eg y \wedge z\) 4. \(x \wedge y \wedge eg z\). All these terms need to be `OR-ed` (\(\vee\)) together, meaning if any of these conditions hold, the whole expression is true.
03

Apply Boolean Law (Distributive Law)

To simplify this, start applying the distributive laws to combine and factor terms where possible. Particularly, check if any terms have common factors across the expressions. In the expressions, we notice the following: Combine using distributive law: \[(x \wedge y \wedge z) \vee (eg x \wedge y \wedge z) = (y \wedge z) \wedge (x \vee eg x)\]. Since \(x \vee eg x\) is a tautology (it is always true), this simplifies to \(y \wedge z\).
04

Continue Simplification

Next, focus on the remaining terms: - Combine \((x \wedge eg y \wedge z)\) with the already simplified part \((y \wedge z)\), check commonalities with \((x \wedge y \wedge eg z)\): We can factor \(x \wedge z\) from both: \[(x \wedge eg y \wedge z) \vee (x \wedge y \wedge z) = x \wedge z \wedge (eg y \vee y)\]. \(eg y \vee y\) is a tautology. This simplifies to \(x \wedge z\).
05

Finish the Simplification

Finally, put together all simplified components: Thus, \((x \wedge y \wedge eg z) \vee (x \wedge y \wedge z) = x \wedge y\). So, the given expression simplifies as follows: - \(y \wedge z\) from Step 3 - \(x \wedge z\) from Step 4 - \(x \wedge y\) from this step.
06

Final Simplification

Combine and write the final expression. We have: \((y \wedge z) \vee (x \wedge z) \vee (x \wedge y)\), which matches the target expression. Since any combination of two variables covers all the constraints, the final expression is correct.

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Key Concepts

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

Combinatorial Network
A combinatorial network in this context is a logical structure used to manipulate or process digital information. Typically, it's made up of primitive logic gates such as AND, OR, and NOT. These gates are combined to evaluate Boolean expressions. By configuring these gates in different ways, we can create circuits that perform specific logical operations. Combinatorial networks are widely used in digital systems like computers and calculators to perform arithmetic and decision-making operations.

Understanding the original problem involves recognizing this network as a series of logic operations combined together. The goal often is to transform these combined operations into a simpler form, achieving the same logical result with potentially fewer gates. This simplification reduces complexity and resource usage.

Identifying the structure of the combinatorial network required to represent the given expression helps in understanding the overall logic design. Being able to simplify complex networks is a fundamental skill in digital electronics.
Simplification
Simplification of a Boolean expression strives to reduce the complexity of the logical network while retaining full functionality. By applying rules and laws of Boolean algebra, a Boolean expression can often be transformed into an equivalent, but simpler form.

To undertake simplification, we aim to minimize the number of terms and operations. This is typically accomplished by employing tautologies, contradictions, and distributive, associative, commutative, identity, and absorption laws.

In our exercise, simplification allows the complex initial expression to be reduced to a more concise form. The essence of simplification here is not losing any logical condition while reducing the resource-intensive use of gates. Understanding and applying simplification reduces redundancy and leads to more efficient computations.
Distributive Law
The distributive law is a critical rule in Boolean algebra, often used for simplifying expressions by factoring and combining terms. It states that for any Boolean variables, the expression \[ A \wedge (B \vee C) = (A \wedge B) \vee (A \wedge C) \] and vice versa is valid.

In the context of simplifying the original problem, the distributive law helps in recognizing common terms across different parts of an expression. For example, pulling out \\( y \wedge z \) from \\((x \wedge y \wedge z) \vee (eg x \wedge y \wedge z)\)can simplify the complex expression substantially.

The law allows restructuring of expressions such that similar terms are grouped, reducing the logical implications needed to evaluate the expression. This restructuring often leads to using fewer logical gates, resulting in a straightforward, cost-effective, and power-efficient circuit design.
Boolean Expression
A Boolean expression is an algebraic representation of a logic circuit composed of binary variables and operations such as AND (\(\wedge\)), OR (\(\vee\)), and NOT (\(eg\)). These expressions are pivotal in digital electronics because they represent the control logic of digital circuits.

Understanding a Boolean expression requires recognizing the binary nature of the variables involved, where each variable can be either true or false. The formulae are utilized to define expressions representing complex logic needed in real-world applications like computing.

The original exercise involves a given Boolean expression that is rather complex in its initial form. By simplifying it to a shorter Boolean expression, we achieve improved efficiency as fewer logic gates might be needed to implement it in a circuit. The more concise an expression, the lesser the potential error rate and resource usage, important for efficient digital circuit design.

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Most popular questions from this chapter

For the following formulas, let the universe be \(\mathbb{R}\). Translate each of the following sentences into a formula (using quantifiers): (a) There is a smallest number. (b) Every positive number has a square root. (Do not use the square root symbol; use only multiplication.) (c) Every positive number has a positive square root. (Again, do not use the square root symbol; use only multiplication.)

Let \(\phi=(p \rightarrow q) \rightarrow((r \wedge \neg s) \rightarrow q)\). For each of the following interpretations of \(p, q, r,\) and \(s,\) compute \(I(\phi)\) using the truth tables for \(\neg, v, \wedge, \rightarrow,\) and \(\leftrightarrow\) (a) \(I(p)=T, I(q)=T, I(r)=F,\) and \(I(s)=T\) (b) \(I(p)=T, I(q)=F, I(r)=T,\) and \(I(s)=F\) (c) \(I(p)=F, I(q)=T, I(r)=T,\) and \(I(s)=F\) (d) \(I(p)=F, I(q)=F, I(r)=T,\) and \(I(s)=F\)

Let \(\phi=(\neg(p \wedge q)) \leftrightarrow(\neg r \vee \neg s)\). For each of the following interpretations of \(p, q, r,\) and \(s,\) compute \(I(\phi)\) using the truth tables for \(\neg, \vee, \wedge, \rightarrow,\) and \(\leftrightarrow\) (a) \(I(p)=T, I(q)=T, I(r)=F,\) and \(I(s)=T\) (b) \(I(p)=T, I(q)=F, I(r)=F,\) and \(I(s)=F\) (c) \(I(p)=F, I(q)=T, I(r)=F,\) and \(I(s)=T\) (d) \(I(p)=F, I(q)=F, I(r)=F,\) and \(I(s)=T\)

Write pseudocode for a program that, given a formula \(\phi,\) finds (i) a logically equivalent formula \(\phi^{\prime}\) in CNF and (ii) a logically equivalent formula \(\phi^{\prime \prime}\) in DNF. The algorithm should be recursive (similar to an induction on formulas) and should not involve the construction of truth tables. Prove the algorithm works. This gives an alternate proof of the theorem that every formula is equivalent to a formula in CNF.

Describe in words the difference between \(\emptyset\) and \(19 .\)

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