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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) \(3 x[(p(x) \vee q(x)) \rightarrow r(x)]\)

Short Answer

Expert verified
The negated and simplified expressions are: a) \( \forall x[\neg p(x) \wedge \neg q(x)] \) b) \( \exists x[\neg p(x) \vee q(x)] \) c) \( \exists x[p(x) \wedge \neg q(x)] \) d) \( \exists x[(p(x) \vee q(x)) \wedge \neg r(x)] \)

Step by step solution

01

- Negate and Simplify the first expression

We have \( \exists x[p(x) \vee q(x)] \). The negation of this statement is \( \neg \exists x[p(x) \vee q(x)] \), which is equivalent to \( \forall x \neg [p(x) \vee q(x)] \). Then, by De Morgan’s Law, \( \forall x \neg [p(x) \vee q(x)] \) is equivalent to \( \forall x[\neg p(x) \wedge \neg q(x)] \).
02

- Negate and Simplify the second expression

The second statement to negate is \( \forall x[p(x) \wedge \neg q(x)] \). The negation of this statement is \( \neg \forall x[p(x) \wedge \neg q(x)] \) which is equivalent to \( \exists x \neg [p(x) \wedge \neg q(x)] \). Then, by De Morgan’s Law, \( \exists x \neg [p(x) \wedge \neg q(x)] \) is equivalent to \( \exists x[\neg p(x) \vee q(x)] \).
03

- Negate and Simplify the third expression

The third statement to negate is \( \forall x[p(x) \rightarrow q(x)] \). The negation gives \( \neg \forall x[p(x) \rightarrow q(x)] \) which is equivalent to \( \exists x \neg [p(x) \rightarrow q(x)] \). According to the definition of material implication, this is equivalent to \( \exists x[p(x) \wedge \neg q(x)] \).
04

- Negate and Simplify the fourth expression

The fourth statement to negate is \( \forall x[(p(x) \vee q(x)) \rightarrow r(x)] \). The negation gives \( \neg \forall x[(p(x) \vee q(x)) \rightarrow r(x)] \) which is equivalent to \( \exists x \neg [(p(x) \vee q(x)) \rightarrow r(x)] \). According to the definition of material implication, this is equivalent to \( \exists x[(p(x) \vee q(x)) \wedge \neg r(x)] \).

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Key Concepts

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

Negation in Logic
Negation in logic is a fundamental concept that involves reversing the truth value of a given statement. If a statement is true, its negation is false, and vice versa. Understanding negation helps us to analyze complex logical expressions by challenging their truths. In predicate logic, to negate quantified statements, we change their quantifiers:
  • For existential quantifiers \(\exists\), we'll switch to universal quantifiers \(\forall\), and
  • For universal quantifiers \(\forall\), we'll switch to existential quantifiers \(\exists\).
Consider the expression \(\exists x[p(x) \vee q(x)]\). Its negation becomes \(\forall x eg [p(x) \vee q(x)]\). This negated form suggests that for all \(x\), it is not the case that \(p(x)\) or \(q(x)\) is satisfied. Negation can be applied to varying complexities of statements and is crucial for constructing proofs or simplifying expressions within predicate logic.
De Morgan's Laws
De Morgan's Laws provide a useful way to simplify the negation of logical expressions that involve conjunctions (and) and disjunctions (or). These laws state:
  • \(eg (A \vee B)\) is equivalent to \(eg A \wedge eg B\)
  • \(eg (A \wedge B)\) is equivalent to \(eg A \vee eg B\)
When applying these laws to predicate logic, they become particularly handy. After negating a statement, De Morgan’s laws help distribute the negation across the components, transforming the internal logical connectors. For instance, in the expression \(\forall x eg [p(x) \vee q(x)]\), De Morgan's law allows us to rewrite this as \(\forall x[eg p(x) \wedge eg q(x)]\). Similarly, for a statement like \(\exists x eg [p(x) \wedge eg q(x)]\), it can be simplified to \(\exists x[eg p(x) \vee q(x)]\). By transforming these expressions, De Morgan's Laws help us understand the structure and implications of the original expressions more clearly.
Material Implication
Material implication is a cornerstone concept in logic, often symbolized as \(\rightarrow\). It suggests that if a statement \(p\) is true, then another statement \(q\) must be true, encapsulated by \(p \rightarrow q\). Another way to view material implication is to consider its negation. The expression \(eg [p(x) \rightarrow q(x)]\) is equivalent to \(p(x) \wedge eg q(x)\). This emphasizes the scenario where \(p(x)\) is true, but \(q(x)\) is false. Understanding material implication is crucial when negating implications: when we negate \(\forall x[p(x) \rightarrow q(x)]\), convert it to \(\exists x eg [p(x) \rightarrow q(x)]\), which further simplifies to \(\exists x[p(x) \wedge eg q(x)]\). The importance of material implication lies in its ability to connect and differentiate between conditions under specific circumstances, proving invaluable in logical reasoning and proofs.

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Most popular questions from this chapter

\begin{aligned} &\text { Let } p, q, r, s \text { denote the following statements: } p: 1 \text { finish writing my computer program } \\ &\text { before lunch; } q \text { : I shall play tennis in the afternoon; } r: \text { The sun is shining; } s \text { The } \\ &\text { humidity is low. Write the following in symbolic form. } \\ &\text { a) If the sun is shining, I shall play tennis this afternoon. } \\ &\text { b) Finishing the writing of my computer program before lunch is necessary for my } \\ &\text { playing tennis this afternoon. } \\ &\text { e) Low humidity and sunshine are sufficient for me to play tennis this afternoon. } \end{aligned}

Construct the truth table for $$ p \leftrightarrow[(q \wedge r) \rightarrow \neg(s \vee r)] . $$

a) Let \(p(x), q(x)\) be open statements in the variable \(x\), with a given universe. Prove that \(\forall x p(x) \vee \forall x q(x) \Rightarrow \forall x[p(x) \vee q(x)] .\) [That is, prove that when the statement \(\forall x p(x) \vee \forall x q(x)\) is true, then the statement \(\forall x[p(x) \vee q(x)]\) is true.] b) Find a counterexample for the converse in part (a). That is, find open statements \(p(x)\), \(q(x)\) and a universe such that \(\forall x[p(x) \vee q(x)]\) is true, while \(\forall x p(x) \vee \forall x q(x)\) is false.

Write the negation of each of the following true statements. For parts (a), (b), and (c) the universe consists of all integers; for parts (d) and (e) the universe comprises all real numbers. a) For all integers \(n\), if \(n\) is not (exactly) divisible by 2 , then \(n\) is odd. b) If the square of an integer is odd, then the integer is odd. c) If \(k, m, n\) are any integers where \(k-m\) and \(m-n\) are odd, then \(k-n\) is even. d) If \(x\) is a real number where \(x^{2}>16\), then \(x<-4\) or \(x>4\). e) For all real numbers \(x\), if \(|x-3|<7\), then \(-4

For each of the following statements provide an indirect proof [as in part (2) of Theorem 2.4] by stating and proving the contrapositive of the given statement. a) For all integers \(k\) and \(l\), if \(k l\) is odd, then \(k, l\) are both odd. b) For all integers \(k\) and \(l\), if \(k+l\) is even, then \(k\) and \(l\) are both even or both odd.

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