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18\. Give the reasons for each step in the following simplifications of compound statements. a) $$ \begin{aligned} &[(p \vee q) \wedge(p \vee \neg q)] \vee q \quad \text { Reasons } \\ & \Leftrightarrow[p \vee(q \wedge \neg q)] \vee q \\ \Leftrightarrow &\left(p \vee F_{0}\right) \vee q \\ \Leftrightarrow & p \vee q \end{aligned} $$

Short Answer

Expert verified
The simplification of the given compound statement mainly involves the use of the distributive law in the first step to achieve \(p \vee (q \wedge \neg q)\). Then, the law of contradiction is used to simplify \(q \wedge \neg q\) to \(F_0\) in the second step, and finally, further simplification yields the final simplified statement \(p \vee q\).

Step by step solution

01

Application of distributive law

The distributive law states that \(A \wedge (B \vee C)\) is equivalent to \((A \wedge B) \vee (A \wedge C)\), and the opposite is also true. The complex compound statement \((p \vee q) \wedge (p \vee \neg q)\) can be simplified using this rule, deriving as a result \(p \vee (q \wedge \neg q)\).
02

Using the law of contradiction

According to the law of contradiction, a proposition and its negation (e.g., 'p and not p') cannot be true simultaneously. So, applying this law to the term \(q \wedge \neg q\) simplifies it to False, written as \(F_0\).
03

Further simplification

The resulting statement after step 2 becomes \((p \vee F_0) \vee q\). Remember, a term OR'ed with a False gives the term itself. So, \(p \vee F_0\) simplifies to p, and having 'p' OR'ed with 'q' gives the final simplified statement \(p \vee q\).

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Key Concepts

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

Distributive Law
Understanding the distributive law in logic is akin to learning a formula that lets us break down complex expressions into simpler parts, not unlike how we can distribute multiplication over addition in arithmetic. In logical operations, it pertains to the relationship between conjunctions (AND, represented by \(\wedge\)), disjunctions (OR, represented by \(\vee\)), and how they can be rearranged without changing the truth value of the statement.

For instance, the distributive law allows us to transform the expression \(A\wedge(B\vee C)\) to \( (A\wedge B)\vee(A\wedge C)\). This reorganization can significantly simplify complex compound statements, like in our exercise where \( (p\vee q)\wedge(p\veeeg q)\) is reinterpreted as \(p\vee(q\wedgeeg q)\), priming us for further simplification.

It's crucial to note that, while the distributive law is intuitive in arithmetic, its use in logic requires us to be vigilant about the context of the problem and the nature of logical operations, ensuring we apply the law correctly to arrive at a valid conclusion.
Law of Contradiction
The law of contradiction is a fundamental principle that asserts a specific restriction: a statement and its negation cannot both be true simultaneously. It's the backbone of logical consistency, preventing contradictory statements from wreaking havoc in our reasonings, such as claiming 'it is raining' and 'it is not raining' at the same time in the same context.

In logical operations, this law allows us to immediately identify certain compound statements as false. For example, \(q\wedgeeg q\) represents a contradiction because it insists that 'q' is both true and not true at the same time, which is impossible. Thus, simplifying this compound statement using the law of contradiction, we get \(F_0\), or simply False. Deploying this law is essential for streamlining complex logical expressions and achieving clarity in mathematical arguments.
Logical Operations in Mathematics
Logical operations in mathematics serve as the building blocks for constructing and interpreting mathematical arguments. They are the tools enabling us to conduct reasoning, set up proofs, and make precise deductions. With operations like 'AND' (\wedge), 'OR' (\vee), 'NOT' (eg), and others, we can construct complex compound statements and manipulate them to reveal underlying truths about the logical structures we're examining.

In the course of solving logical problems, we often employ these operations to simplify statements or prove equivalencies. For instance, in our original exercise, steps of logical operations transformed the compound statement through application of distributive law and the law of contradiction to arrive at the simplest form \(p\vee q\). This simplified statement maintains the truth values of the original, yet it's more straightforward and easier to apply in further analyses or arguments. Understanding how to effectively use logical operations is key to success in mathematics and other disciplines that rely on rigorous reasoning.

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Most popular questions from this chapter

17\. For any statements \(p, q\), prove that a) \(\neg(p \downarrow q) \Leftrightarrow(\neg p \uparrow \neg q)\) b) \(\neg(p \uparrow q) \Leftrightarrow(\neg p \downarrow \neg q)\)

18\. Negate and simplify each of the following. a) \(\exists x[p(x) \vee q(x)]\) b) \(\forall x[p(x) \wedge \neg q(x)]\) c) \(\forall x[p(x) \rightarrow q(x)]\) d) \(\exists x[(p(x) \vee q(x)) \rightarrow r(x)]\)

23\. In the arithmetic of real numbers, there is a real number, namely 0, called the identity of addition because \(a+0=\) \(0+a=a\) for cvery real number \(a\). This may be expressed in symbolic form by $$ \exists z \forall a[a+z=z+a=a] $$ (We agree that the universe comprises all real numbers.) a) In conjunction with the existence of an additive identity is the existence of additive inverses. Write a quantified statement that cxpresses "Every real number has an additive inverse." (Do not use the minus sign anywhere in your statement.) b) Write a quantified statement dealing with the existence of a multiplicative identity for the arithmetic of real numbers. c) Write a quantified statement covering the existence of multiplicative inverses for the nonzero real numbers. (Do not use the exponent \(-1\) anywhere in your statement.) d) Do the results in parts (b) and (c) change in any way when the universe is restricted to the integers?

4\. Let \(n\) be a positive integer greater than 1 . We call \(n\) prime if the only positive integers that (exactly) divide \(n\) are 1 and \(n\) itself. For example, the first seven primes are \(2,3,5,7,11\), 13 , and 17. (We shall learn more about primes in Chapter 4.) Use the method of exhaustion to show that every integer in the universe \(4,6,8, \ldots, 36,38\) can be written as the sum of two primes.

4\. For each of the following pairs of statements, use Modus 'onens or Modus Tollens to fill in the blank line so that a valid rgument is presented. a) If Janice has trouble starting her car, then her daughter Angela will check Janice's spark plugs. Janice had trouble starting her car. b) If Brady solved the first problem correctly, then the answer he obtained is 137 . Brady's answer to the first problem is not 137 . c) If this is a repeat-until loop, then the body of this loop is executed at least once. The body of the loop is executed at least once. d) If Tim plays basketball in the afternoon, then he will not watch television in the evening.

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