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Chapter 2: Problem 7

a) For primitive statements \(p, q\), find the dual of the statement \((\neg p \wedge-q) \vee\left(T_{0} \wedge p\right) \vee p\). b) Use the laws of logic to show that your result from part (a) is logically equivalent to \(p \wedge \neg q\).

Short Answer

Expert verified
The dual of the statement \((\neg p \land -q) \lor (T_0 \land p) \lor p\) is \(F\). However, this result is not logically equivalent to \(p \land \neg q\).

Step by step solution

01

Find the dual of the statement

The dual statement of \((\neg p \wedge \neg q) \lor (T_0 \land p) \lor p\) can be obtained by interchanging \(\land\) with \(\lor\) and vice versa. Also, \(T_0\) should be transformed into \(\neg T_0\). So, the dual of the statement becomes \((\neg p \lor \neg q) \land (\neg T_0 \lor p) \land \neg p\). The terms can be rearranged to get \((\neg p \lor p) \land (\neg p \lor \neg q) \land \neg T_0\).
02

Simplify the parts of the expression

Simplify \((\neg p \lor p)\) to \(T\) and \(\neg T_0\) to \(F\). This results in the simplified dual statement getting \(T \land (\neg p \lor \neg q) \land F\).
03

Simplifying the dual statement using laws of logical equivalence

Using the law of identity which states \(p \land T \equiv p\) and \(p \land F \equiv F\), the expression will simplify to \(F\). This is because when there is an AND operation with False (F), the result is always False (F).
04

Logical equivalence of the result from part (a) to \(p \land \neg q\)

The given logic equivalence to prove is \(F \equiv p \land \neg q\). According to the laws of logic, it is not possible for the statement to be true, hence the result is not logically equivalent to \(p \land \neg q\).

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Key Concepts

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

Dual of a Statement
In logic, the concept of the "dual of a statement" refers to a systematic method of transforming a logical statement. When creating a dual, we swap conjunctions (\(\land\)) with disjunctions (\(\lor\)) and vice versa. Additionally, logical constants like True (\(T\)) and False (\(F\)) are reversed. So, \(T\) becomes \(F\) and \(F\) becomes \(T\).To better understand, consider the proposition \((eg p \land eg q) \lor (T_0 \land p) \lor p\). To find its dual, swap all \(\land\) with \(\lor\) and replace all \(T_0\) with \(eg T_0\). The result is \((eg p \lor eg q) \land (eg T_0 \lor p) \land eg p\). Rearranging gives us \((eg p \lor p) \land (eg p \lor eg q) \land eg T_0\). This transformation is essential in determining logical equivalences between statements.
Laws of Logic
The "laws of logic" are fundamental principles that help simplify and prove logical statements. Key laws include identity, domination, idempotent, and double negation, among others. They are vital tools in verifying logical equivalence between statements.For instance, by applying these laws, we can simplify logical expressions. In our exercise, \((eg p \lor p) \land (eg p \lor eg q) \land eg T_0\) based on the laws simplifies as follows:
  • Using the law of \((eg p \lor p) \equiv T\) (tautology), \((eg p \lor p)\) simplifies to \(T\).
  • The law of identity \((p \land T) \equiv p)\) helps further simplify.
  • Additionally, since anything ANDed with \(F\) results in \(F\), the statement simplifies to \(F\), which was shown not to be logically equivalent to \(p \land eg q\).
These logical laws not only simplify expressions but also validate the equivalency or discrepancy between logical propositions.
Primitive Statements
"Primitive statements" are the simplest forms of propositions. These are basic statements which do not include logical connectors like \(\land\), \(\lor\), or \(eg\). In terms of the components, examples typically include symbols like \(p\) or \(q\).Understanding primitive statements is crucial as they form the building blocks of more complex logical expressions. In our given exercise, \(p\) and \(q\) are primitive statements used within the logical structure. Through operations and transformations, these primitive statements evolve into compound statements that convey different meanings or equivalences based on how they are manipulated using logical laws and duals.In exploring logical equivalences or in calculating duals, starting with a firm grasp of these primitive components ensures accuracy as you progress to more complex derivations.

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Most popular questions from this chapter

\text { Prove that for every integer } n, n^{2} \text { is even if and only if } n \text { is even. }

1\. Let \(p(x), q(x)\) denote the following open statements. $$ p(x): x \leq 3 \quad q(x): x+1 \text { is odd } $$ If the universe consists of all integers, what are the truth values of the following state ments? a) \(p(1)\) b) \(q(1)\) c) \(\neg p(3)\) d) \(q(6)\) e) \(p(7) \vee q(7)\) f) \(p(3) \wedge q(4)\) g) \(p(4)\) b) \(\neg(p(-4) \vee q(-3))\) i) \(\rightarrow p(-4) \wedge \neg q(-3)\)

Let \(p(x, y), q(x, y)\) denote the following open statements: $$ p(x, y): x^{2} \geq y \quad q(x, y): x+2

In calculus the definition of the limit \(L\) of a sequence of real numbers \(r_{1}, r_{2}, r_{3}, \ldots\) can be given as $$ \lim _{n \rightarrow \infty} r_{n}=L $$ if (and only if) for every \(\epsilon>0\) there exists a positive integer \(k\) so that for all integers \(n\), if \(n>k\) then \(\left|r_{n}-L\right|<\epsilon\). In symbolic form this can be expressed as $$ \lim _{n \rightarrow \infty} r_{n}=L \Leftrightarrow \forall e>0 \quad \exists k>0 \quad \forall n\left[(n>k) \rightarrow\left|r_{n}-L\right|<\epsilon\right] . $$ Express \(\lim _{n \rightarrow \infty} r_{n} \neq L\) in symbolic form.

Determine whether each of the following is true or false. Here \(p, q\), and \(r\) are arbitrary statements. a) An equivalent way to express the converse of " \(p\) is sufficient for \(q^{* \prime}\) is " \(p\) is necessary for \(q\), b) An equivalent way to express the inverse of " \(p\) is necessary for \(q\) " is " \(7 q\) is sufficient for \(\neg p . "\) c) An equivalent way to express the contrapositive of " \(p\) is necessary for \(q^{\prime \prime}\) is " \(\neg q\) is necessary for \(\neg p .^{31}\) d) An equivalent way to express the converse of \(p \rightarrow(q \rightarrow r)\) is \((\neg q \vee r) \rightarrow p\).
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