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Show that the following formulas from Table 2.5 are tautologies: (a) \((p \wedge p) \leftrightarrow p\) (b) \((p \wedge(p \rightarrow q)) \rightarrow q\) (c) \((p \rightarrow r) \leftrightarrow(\neg r \rightarrow \neg p)\)

Short Answer

Expert verified
All three formulas are tautologies because they are true for any truth values of the variables.

Step by step solution

01

Define a Tautology

A formula is called a tautology if it is true for all possible truth values of its variables. This means, regardless of how you assign values to the variables within the formula (true or false), the outcome will always be true.
02

Analyze Formula (a)

Formula (a) is \((p \wedge p) \leftrightarrow p\). We start by examining the term \(p \wedge p\), which simplifies to \(p\) since \(p \wedge p\) is true if and only if \(p\) is true. So, the formula becomes \(p \leftrightarrow p\), and this expression is always true because any statement is always equivalent to itself.
03

Examine Formula (b)

Consider \((p \wedge(p \rightarrow q)) \rightarrow q\). First, \(p \rightarrow q\) is true unless \(p\) is true and \(q\) is false. Thus \(p \wedge(p \rightarrow q)\) means \(p\) is true, and \(p \rightarrow q\) is also true, implying \(q\) must be true. Then the implication \((...) \rightarrow q\) is always true because if the premise is true, so must be \(q\). If the premise is false, the implication is automatically true by definition.
04

Explore Formula (c)

Consider \((p \rightarrow r) \leftrightarrow (eg r \rightarrow eg p)\). By using the contra-positive law, \(p \rightarrow r\) is logically equivalent to \(eg r \rightarrow eg p\). Hence, the formula \((p \rightarrow r) \leftrightarrow (eg r \rightarrow eg p)\) is always true as it expresses a logical equivalence.
05

Conclude

After evaluating each formula, we see that they hold true for all possible truth values of \(p\), \(q\), and \(r\). Therefore, each formula is a tautology, by definition.

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Key Concepts

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

Logical Equivalence
Logical equivalence is a fundamental concept in propositional logic. It occurs when two statements or propositions have the same truth value in every possible scenario. When we say that two formulas are logically equivalent, we mean that no matter how you assign truth values to their variables, the truth values of the entire expressions will always match in every case. For example, consider this formula: - \( (p \rightarrow r) \leftrightarrow (eg r \rightarrow eg p) \). Using the rule of contraposition, we know that - \( p \rightarrow r \) is equivalent to - \( eg r \rightarrow eg p \). Thus, the expression is always true, highlighting logical equivalence. Logical equivalence is vital for understanding tautologies because tautologies often arise from logically equivalent expressions.
Propositional Logic
Propositional logic is a branch of logic that deals with propositions, which are declarative sentences that can either be true or false. This field of logic uses symbols to represent logical forms, which allows us to analyze statements systematically. The use of operators such as "and" (\(\wedge\)), "or" (\(\vee\)), and "implies" (\(\rightarrow\)) are key in building complex expressions. In propositional logic, we often look at formulas and determine their truthfulness based on different combinations of truth values for the variables involved. This ability to apply rules and derive new statements from existing ones makes propositional logic a powerful tool for reasoning and problem solving. In exercises like this, understanding propositional logic helps classify formulas as tautologies, contradictions, or contingencies, guiding us in logical reasoning.
Truth Tables
Truth tables are an essential tool in propositional logic, providing a clear way to visualize how the truth values of variables affect the outcome of logical expressions. Truth tables list all possible combinations of truth values for their input variables, and then calculate the resulting truth value for the expression. This process helps us confirm whether a formula is always true, sometimes true, or never true. For instance, when examining a formula to verify it as a tautology, a truth table allows us to - systematically check that the formula holds true across all possible truth value assignments. This method ensures that there are no exceptions where the formula becomes false, confirming it as a tautology. By using truth tables, anyone can become more adept at understanding and verifying logical expressions systematically.

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Most popular questions from this chapter

Let \(\phi=(p \vee q) \rightarrow(r \wedge \neg s)\). 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) \(I(p)=T, I(q)=T, I(r)=T,\) and \(I(s)=F\) (b) \(I(p)=T, I(q)=T, I(r)=F,\) and \(I(s)=F\) (c) \(I(p)=F, I(q)=T, I(r)=T,\) and \(I(s)=T\) (d) \(I(p)=F_{,} I(q)=F_{,} I(r)=T,\) and \(I(s)=T\)

Find formulas equivalent to the following formulas with all the negations "pushed inward to the proposition letters": (a) \(\neg(p \wedge T)\) (b) \(((p \rightarrow q) \rightarrow r) \rightarrow F\) (c) \(((p \rightarrow q) \rightarrow r) \rightarrow T\) (d) \((p \leftrightarrow q) \leftrightarrow r\) (c) \((p \leftrightarrow q) \leftrightarrow F\) (Hint: Look for a way to simplify this last one.) (Note: The method given to "push negations inward" does not always give the shortest formula that is equivalent to the given formula and has \(\neg\) applied only to proposition letters.)

(a) Show that every formula containing only \(k\) (different) proposition letters is equivalent to a \(k-\mathrm{DNF}\) formula. (b) Show that \(p \leftrightarrow q\) is not equivalent to any 1 -DNF formula. (c) Show that for every natural number \(k\) (including 0 ), there is a formula containing only \(k+1\) (different) proposition letters that is not equivalent to any \(k-D N F\) formula.

(a) Find the resolvant of \((p \vee q)\) and \((\neg p \vee r)\) on \(p\). (b) Find the resolvant of \((p \vee q \vee r \vee s)\) and \((\neg p \vee \neg q \vee t)\) on \(p\). (c) Find the resolvant of \((p \vee q)\) and \(\neg p\) on \(p\). (d) Find the resolvant of \((p)\) and \((\neg p)\) on \(p\). (e) Which resolvant above from parts (a) through (d) is a tautology? Which is tautologically false?

(a) The conjunction of \(n\) formulas \(p_{1}, p_{2}, \ldots, p_{n}\) is defined to be the formula \(\left(\ldots\left(\left(p_{1} \wedge p_{2}\right) \wedge p_{3}\right) \wedge \ldots\right) \wedge p_{n} .\) For \(n=0,\) there is a special case: The conjunction of zero formulas is defined to be \(T\). For \(n=1\), that conjunction simplifies to \(p_{1}\). Let \(\phi\) be the conjunction of \(p_{1}, p_{2}, \ldots, p_{n} .\) Prove that for any interpretation \(I, I(\phi)=T\) if and only if \(I\left(p_{i}\right)=T\) for each \(i\) such that \(1 \leq i \leq n .\) (Hint: Use induction.) (b) Let \(\phi\) be the formula $$ \left(\ldots\left(\left(p_{1} \leftrightarrow p_{2}\right) \leftrightarrow p_{3}\right) \leftrightarrow \ldots\right) \leftrightarrow p_{n} $$ for \(n \geq 1 .\) For what interpretations \(I\) is \(I(\phi)=T ?\) (Hint: The answer involves counting how many of the \(p_{i}\) 's are true in \(I\). Prove the result by induction on \(n\).)

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