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(a) Show that \((p \vee q)\) is an alphabetic variant of \((q \vee p)\). (b) Show that the relation of being an alphabetic variant is an equivalence relation. (c) Show that if \(\psi\) is an alphabetic variant of \(\phi .\) then \(\phi\) is a tautology (respectively, is satisfiable, is unsatisfiable) if and only if \(\psi\) is a tautology (respectively, is satisfiable, is unsatisfiable). (d) Show that \(\phi\) being an alphabetic variant of \(\psi\) does not imply that \(\phi\) and \(\psi\) are tautologically equivalent.

Short Answer

Expert verified
(a) Yes, they are alphabetic variants. (b) The relation is an equivalence relation. (c) Alphabetic variants preserve tautology and satisfiability. (d) Alphabetic variants don't ensure tautological equivalence.

Step by step solution

01

Definitions and Alphabetic Variation

To solve these questions, we first need to clarify what an alphabetic variant and an equivalence relation mean. An *alphabetic variant* refers to a logical formula where the variables were renamed, but the logical structure remains unaltered. An *equivalence relation* is a particular kind of relation, which is reflexive, symmetric, and transitive.
02

Alphabetic Variant Example

(a) To show that \((p \vee q)\) is an alphabetic variant of \((q \vee p)\), observe that due to the commutative property of disjunction (\(\vee\)), rearranging the variables doesn't change the logical meaning. Therefore, \((p \vee q)\) is indeed an alphabetic variant of \((q \vee p)\) because they are identical up to ordering and renaming.
03

Prove Equivalence Relation Properties

(b) The relation of being an alphabetic variant is shown to be an equivalence relation through: 1. **Reflexivity**: Any formula \(\phi\) is trivially an alphabetic variant of itself.2. **Symmetry**: If \(\phi\) is an alphabetic variant of \(\psi\), then \(\psi\) is an alphabetic variant of \(\phi\), due to the nature of renaming.3. **Transitivity**: If \(\phi\) is an alphabetic variant of \(\psi\), and \(\psi\) is an alphabetic variant of \(\chi\), then \(\phi\) is an alphabetic variant of \(\chi\). This is because both transformations preserve logical structure.
04

Tautology and Satisfiability

(c) Consider that if \(\psi\) is an alphabetic variant of \(\phi\), the logical structure remains the same under variable renaming, so truth values do not change. Thus, if \(\phi\) is a tautology, \(\psi\) must also be a tautology, and similarly for satisfiability and unsatisfiability.
05

Tautological Equivalence

(d) Being alphabetic variants doesn’t imply tautological equivalence. Consider \(\phi = (p \wedge eg p)\) and \(\psi = (q \wedge eg q)\); both are unsatisfiable regardless of variable naming. However, in cases where variable order or logical operators differ, tautological equivalence might not hold despite alphabetic variance.

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Key Concepts

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

Equivalence Relation
Before we dive into the intricacies of equivalence relations in logical formulas, it's crucial to understand what an equivalence relation is. An equivalence relation needs to have three properties: reflexivity, symmetry, and transitivity.
  • **Reflexivity** ensures that any logical formula is related to itself. For instance, any logical expression \( \phi \) is always an alphabetic variant of itself.
  • **Symmetry** means if one formula is an alphabetic variant of another, then the opposite is also true; that is, they can be swapped without affecting the relation.
  • **Transitivity** guarantees that if you have one variant relationship between two formulas and another with a third one, you can connect the first and third directly.
These components confirm that the relationship between alphabetic variants is indeed an equivalence relation. This understanding is foundational to exploring how different logical formulas can share equivalent logical structures despite differences in literals or variable names.
Tautology
A tautology in logic is a formula that remains true in every possible interpretation, irrespective of the truth values of the variables it contains. For example, the statement \( (p \lor eg p) \) is a classic tautology.
What makes tautological analysis intriguing is that any logical formula which is a tautology maintains its truth status even if the variable names change—or if it is an alphabetic variant. This property reflects the consistency and redundancy aspects in propositional logic.

In simpler terms, if you transform a formula into a variant by renaming variables or altering the configuration (without changing the logical operators), and if it was a tautology before, it remains one. This characteristic enables us to assess the intrinsic 'truth' nature of logical formulas without being distracted by superficial alterations.
Satisfiability
Satisfiability is a concept in propositional logic that identifies whether there exists an interpretation that makes a logical formula true. If there is at least one configuration of variable assignments that lead the formula to be true, it's considered satisfiable.

When dealing with alphabetic variants, it's important to note that satisfiability is preserved. If a formula \( \phi \) is satisfiable, then any alphabetic variant \( \psi \) would also be satisfiable since the renaming of variables or reordering doesn’t affect the truth assignment possibilities.
This feature of satisfiability provides a robust framework for evaluating logical systems and understanding how underlying structures can be altered without losing their essential characteristics and abilities to "satisfy." It serves as a fundamental tool in various fields, such as computer science and mathematical logic, where evaluating the truth of expressions efficiently is crucial.
Logical Formulas
Logical formulas represent structured expressions consisting of variables, logical operators (like AND, OR, NOT), and possibly constants. They are the backbone of any logical system, providing a means to express conditions and assertive statements.
An alphabetic variant of a logical formula is essentially the same structure but with variable names changed. It is essential to note that the core logical operations and their relationships remain unchanged. Thus, alphabetic variants, although expressed differently, exhibit the same logical characteristics in terms of truth values.
  • Proper handling and understanding of logical formulas ensure clarity in situations like software design, data analysis, and mathematical theorem proving.
  • They help in constructing logical arguments and understanding proofs and propositions systematically.
These formulas often serve as a gateway to more complex systems and are vital for those studying areas that require critical thinking and problem-solving strategies linked with logic fundamentals.

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Most popular questions from this chapter

Simplify the following boolean expressions: (a) \((x \wedge y) \vee(x \wedge \neg y) \vee(\neg x \wedge y) \vee(\neg x \wedge \neg y)\) (b) \((x \wedge y \wedge z) \vee(x \wedge \neg y \wedge z) \vee(\neg x \wedge y \wedge \neg z) \vee(\neg x \wedge \neg y \wedge z)\) (c) \((x \wedge y \wedge \neg z) \vee(x \wedge \neg y \wedge z) \vee(x \wedge \neg y \wedge \neg z)\)

Find a formula in negation normal form equivalent to the negation of $$ \exists x \forall y \forall z(P(x, y, z)) $$.

The country of Ost is inhabited only by people who either always tell the truth or always tell lies and who will respond to questions only with a "yes" or a "no." A tourist comes to a fork in a road, where one branch leads to the capital and the other does not. There is no sign indicating which fork to take, but Mr. Zed, who is a resident of Ost, comes along. What single question should the tourist ask Mr. Zed to determine which fork in the road to take?

(a) Show that the following formula in CNF is unsatisfiable: $$ (p \vee q) \wedge(p \vee \neg q) \wedge(\neg p \vee q) \wedge(\neg p \vee \neg q) $$ (b) Show that the following formula in CNF is unsatisfiable: $$ \begin{array}{c} (p \vee q \vee r) \wedge(p \vee \neg q \vee r) \wedge(\neg p \vee q \vee r) \wedge(\neg p \vee \neg q \vee r) \\ \wedge(p \vee q \vee \neg r) \wedge(p \vee \neg q \vee \neg r) \wedge(\neg p \vee q \vee \neg r) \wedge(\neg p \vee \neg q \vee \neg r) \end{array} $$ Can you find an easier argument than just writing the entire truth table? (c) Generalize the above to some class of CNF formulas on an arbitrary number \(n \geq 1\) of proposition letters, and prove it by induction on \(n\).

The first stage of the method described to "push negations inward" was a method to climinate \(\rightarrow\) 's and \(\leftrightarrow\) 's. Prove that in the method to eliminate them, the process of substituting always stops, Consider, for example, the substitution in the formula $$ (p \leftrightarrow q) \leftrightarrow(r \leftrightarrow s) $$ If the substitution is first performed on the second \(\leftrightarrow\), the resultant formula is $$ ((p \leftrightarrow q) \rightarrow(r \leftrightarrow s)) \wedge((r \leftrightarrow s) \rightarrow(p \leftrightarrow q)) $$ which has more \(\leftrightarrow\) 's to replace than in the original formula! At first sight, one might expect that if the substitutions are made in the wrong order, the process might continue generating more \(\leftrightarrow\) 's at each stage, and the process might continue forever. (Hint: One method is to, instead of just counting the number of \(\leftrightarrow\) symbols, put a weight on each \(\leftrightarrow\) symbol, with the weight of the \(\leftrightarrow\) symbol in \(\psi \leftrightarrow x\) being dependent on the number of \(\leftrightarrow\) 's in \(\psi\) and \(\chi\). If the correct method of calculating weights is used, it can be shown that the total weight of the \(\leftrightarrow\) 's decreases with each substitution.

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