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Show that \((p \vee q) \wedge(\neg p \vee r) \rightarrow(q \vee r)\) is a tautology.

Short Answer

Expert verified
It is a tautology because \( (p \vee q) \wedge (\eg p \vee r) \rightarrow (q \vee r) \) is always true.

Step by step solution

01

Understand the Statement

We need to show that \( (p \vee q) \wedge (\eg p \vee r) \rightarrow (q \vee r) \) is a tautology. A tautology is a statement that is true for all possible truth values of the variables involved.
02

Construct a Truth Table

Create a truth table for all possible values of \( p \), \(q \), and \(r \). There are 8 possible combinations:\[\begin{array}{cccccccc}p & q & r & p \vee q & \eg p & \eg p \vee r & (p \vee q) \wedge (\eg p \vee r) & q \vee r \ \hlineT & T & T & T & F & T & T & T \T & T & F & T & F & F & F & T \T & F & T & T & F & T & T & T \T & F & F & T & F & F & F & F \F & T & T & T & T & T & T & T \F & T & F & T & T & F & F & T \F & F & T & F & T & T & F & T \F & F & F & F & T & T & F & F \end{array}\]
03

Evaluate the Conditional Statement

Look at the columns for \( (p \vee q) \wedge (\eg p \vee r) \) and \( q \vee r \). Since we are evaluating an implication, the statement is true when \( (p \vee q) \wedge (\eg p \vee r) \) is false, and if it is true implies that \( (q \vee r) \) must also be true. By checking the truth table:\[\begin{array}{cccccc}p & q & r & (p \vee q) \wedge(\eg p \vee r) & q \vee r & (p \vee q) \wedge(\eg p \vee r) \rightarrow (q \vee r) \ \hlineT & T & T & T & T & T \T & T & F & F & T & T \T & F & T & T & T & T \T & F & F & F & F & T \F & T & T & T & T & T \F & T & F & F & T & T \F & F & T & F & T & T \F & F & F & F & F & T \end{array}\]
04

Conclusion

Since \( (p \vee q) \wedge (\eg p \vee r) \rightarrow (q \vee r)\) is true for all combinations of truth values of \( p \), \( q \), and \( r \), it is a tautology.

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Key Concepts

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

Propositional Logic
Propositional logic, also known as Boolean logic, is the branch of logic that deals with propositions. A proposition is a statement that can either be true or false, but not both. Examples of propositions include statements like 'It is raining' (which is either true or false). Symbols like 'p' or 'q' are often used to represent propositions. In propositional logic, we use connectives such as:
  • \(\vee\) for 'OR'
  • \(\wedge\) for 'AND'
  • \(eg\) for 'NOT'
  • \(\rightarrow\) for 'IMPLICATION'
These connectives allow us to build complex logical expressions from simple propositions. For example, \(p \vee q\wedge eg r\rightarrow s\) combines several propositions and connectives into a single statement. Understanding how to manipulate these expressions is key to mastering propositional logic.
Truth Table
A truth table is a useful tool in propositional logic. It shows all possible truth values for a given logical expression. For each combination of truth values of variables, the table allows us to see whether the logical expression as a whole is true or false. For example, the truth table for the expression \( (p \vee q) \wedge (eg p \vee r) \rightarrow (q \vee r) \) involves the variables 'p', 'q', and 'r'. By listing all possible values (true or false) for these variables, we can systematically determine the truth value of the entire expression:
1. Start with all possible truth combinations for 'p', 'q', and 'r' (there are 8 combinations).
2. Calculate intermediate results like \( p \vee q \) and \( eg p \vee r \).
3. Combine intermediate results to get the value of \( (p \vee q) \wedge(eg p \vee r) \).
4. Finally, check if \( (p \vee q) \wedge(eg p \vee r) \rightarrow (q \vee r) \) is true for all combinations. A statement is a tautology if it is true for all possible combinations.
Implication
In propositional logic, an implication is expressed using the connective \( \rightarrow \). An implication statement \(p \rightarrow q \) means 'if p then q'. It is true in all situations except when 'p' is true and 'q' is false. To understand this better, let鈥檚 look at an example:
Imagine 'p' stands for 'It is raining' and 'q' stands for 'The ground is wet'. The statement \( p \rightarrow q \) means 'if it is raining, then the ground is wet'. This is intuitively true: whenever it rains, you expect the ground to be wet. In a truth table, the implication \( p \rightarrow q \) is false only when 'p' is true and 'q' is false.
In complex logical expressions, implications help in deriving conclusions. For the expression \( (p \vee q) \wedge(eg p \vee r) \rightarrow (q \vee r) \), the implication ensures the right side ('q' or 'r') must be true whenever the left side (a combination of terms involving 'p', 'q', and 'r') is true. By proving this holds for all truth value combinations, we can conclude the statement is a tautology.

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Most popular questions from this chapter

Use resolution to show that the hypotheses 鈥淚t is not raining or Yvette has her umbrella,鈥 鈥淵vette does not have her umbrella or she does not get wet,鈥 and 鈥淚t is raining or Yvette does not get wet鈥 imply that 鈥淵vette does not get wet.鈥

What is wrong with this argument? Let \(S(x, y)\) be "\(x\) is shorter than \(y\) ." Given the premise \(\exists s S(s, \text { Max })\) , it follows that \(S(\text { Max, Max })\) . Then by existential generalization it follows that \(\exists x S(x, x),\) so that someone is shorter than himself.

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Exercises \(61-64\) are based on questions found in the book Symbolic Logic by Lewis Carroll. Let P(x), Q(x), R(x), and S(x) be the statements 鈥渪 is a duck,鈥 鈥渪 is one of my poultry,鈥 鈥渪 is an officer,鈥 and 鈥渪 is willing to waltz,鈥 respectively. Express each of these statements using quantifiers; logical connectives; and P(x), Q(x), R(x), and S(x). a) No ducks are willing to waltz. b) No officers ever decline to waltz. c) All my poultry are ducks. d) My poultry are not officers. e) Does (d) follow from (a), (b), and (c)? If not, is there a correct conclusion?

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