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Find formulas in DNF equivalent to each of the following formulas, and find at least two interpretations that make each formula satisfiable: (a) \(((p \rightarrow q) \rightarrow r) \rightarrow F\) (b) \(\neg(p \leftrightarrow q) \leftrightarrow r\) (c) \((\neg r) \rightarrow(((p \vee q) \rightarrow r) \rightarrow \neg q)\)

Short Answer

Expert verified
DNF for (a): \((\neg p \vee q) \wedge \neg r\). DNF for (b): complex. DNF for (c): \((r \wedge q) \vee (p \wedge \neg q)\).

Step by step solution

01

Convert part (a) to DNF

The formula \( ((p \rightarrow q) \rightarrow r) \rightarrow F \) is equivalent to \( eg ((p \rightarrow q) \rightarrow r) \). First, translate \( p \rightarrow q \) to \( eg p \vee q \), then \( (eg p \vee q) \rightarrow r \) to \( eg(eg p \vee q) \vee r \). Simplifying this gives \( (p \wedge eg q) \vee r \). The negation \( eg((p \wedge eg q) \vee r) \) is equivalent to \( eg(p \wedge eg q) \wedge eg r \). Further simplification gives \( (eg p \vee q) \wedge eg r \), which is the DNF.
02

Interpretations for part (a)

To satisfy \( (eg p \vee q) \wedge eg r \), both conditions must be true:1. \( eg p \) and \( eg r \), with an interpretation like \( (p = F, q = T, r = F) \).2. \( q \) and \( eg r \), with an interpretation such as \( (p = T, q = T, r = F) \).
03

Convert part (b) to DNF

The formula \( eg(p \leftrightarrow q) \leftrightarrow r \) can be rewritten as \( (p \leftrightarrow q) \oplus r \). First, \( p \leftrightarrow q = (p \wedge q) \vee (eg p \wedge eg q) \). The negation is \( (eg p \wedge q) \vee (p \wedge eg q) \).So, the full expression is \( [((eg p \wedge q) \vee (p \wedge eg q)) \wedge eg r] \vee [((p \wedge q) \vee (eg p \wedge eg q)) \wedge r] \), which is in DNF.
04

Interpretations for part (b)

To satisfy the DNF, here are two interpretations:1. From \((eg p \wedge q) \wedge eg r\): \( (p = F, q = T, r = F) \).2. From \((p \wedge q) \wedge r\): \( (p = T, q = T, r = T) \).
05

Convert part (c) to DNF

The formula \( (eg r) \rightarrow (((p \vee q) \rightarrow r) \rightarrow eg q) \) is first converted to \( r \vee (((p \vee q) \rightarrow r) \rightarrow eg q) \). Translate \( (p \vee q) \rightarrow r \) as \( eg(p \vee q) \vee r \) and simplify to \( (eg p \wedge eg q) \vee r \). The full expression becomes \( r \vee [((eg p \wedge eg q) \vee r) \rightarrow eg q] \). Translate this to \( r \vee eg((eg p \wedge eg q) \vee r) \vee q \), which simplifies to \((r \wedge q) \vee (p \wedge eg q)\), which is in DNF.
06

Interpretations for part (c)

Here are two satisfying interpretations:1. \((r \wedge q)\): \( (p = F, q = T, r = T) \).2. \((p \wedge eg q)\): \( (p = T, q = F, r = F) \).

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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 an important concept when working with logical formulas. Two formulas are logically equivalent if they show the same truth value in every possible interpretation.
For example, if we have a formula like \( (p \rightarrow q) \rightarrow r \), it is logically equivalent to its transformed version in disjunctive normal form (DNF), which will have the same truth table.
Logical equivalence allows us to replace a formula with another form that might be easier to analyze or solve.
  • Logical equivalence helps simplify complex logical expressions.
  • Proving equivalency can reveal insights into the logical structure of formulas.
Understanding logical equivalence is crucial for tasks like converting formulas into DNF, as we can ensure the transformation does not change the logical meaning.
Satisfiability
Satisfiability refers to the assignment of truth values to variables in a logical formula that makes the entire formula true. If such an assignment exists, the formula is called satisfiable; otherwise, it is unsatisfiable.
For example, a formula in DNF can easily show satisfiability. The formula \( (p \wedge eg q) \vee (r \wedge q) \) is satisfiable because at least one of its disjunctions can be true.
Furthermore:
  • Each clause in a DNF should contain variables that allow at least one interpretation to hold true.
  • Analyzing the satisfiability of logical formulas helps in determining their usability in logical operations.
It's a fundamental step to check satisfiability when working with logical formulas, as it determines if there are truth conditions under which the formula holds.
Interpretations
Interpretations are assignments of truth values to the variables in a logical formula. These assignments help determine the truth value of the entire formula.
When discussing interpretations, we are essentially working with the possible states that each variable can take.
For examples from the previous exercises:
  • The formula \( (eg p \vee q) \wedge eg r \) satisfies the interpretation \( (p = F, q = T, r = F) \).
  • Such interpretations provide concrete examples of how formulas behave under different conditions.
By exploring different interpretations, one can find which combinations of truth assignments yield a true or false outcome for the entire formula. This exploration is key for ensuring the consistency and correctness of logical systems.
Logical formulas
Logical formulas are expressions formed by combining variables and logical operators such as \( \wedge \) (and), \( \vee \) (or), \( eg \) (not), \( \rightarrow \) (implies), and \( \leftrightarrow \) (if and only if).
These formulas are used to represent logical statements mathematically.
  • A logical formula like \( eg(p \leftrightarrow q) \leftrightarrow r \) can be broken down and converted into DNF to assess its structure more easily.
  • Logical formulas are the building blocks for constructing complex logical statements and exploring logical arguments.
Understanding logical formulas is essential for creating precise propositions and for applying logical reasoning in mathematics and computer science. They serve as the foundation for more advanced topics like logical equivalence and satisfiability.

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Most popular questions from this chapter

Write the truth tables for the following formulas, Use the truth table to determine whether any of these formulas is a tautology. (a) \(((p \rightarrow q) \wedge(q \rightarrow r)) \rightarrow(p \leftrightarrow r)\) (b) \(((p \rightarrow q) \wedge(q \rightarrow r)) \rightarrow(p \rightarrow r)\) (c) \(((p \rightarrow q) \rightarrow r) \rightarrow(p \rightarrow(q \rightarrow r))\) (d) \((p \rightarrow(r \vee q)) \rightarrow((p \rightarrow r) \vee(p \rightarrow q))\) (e) \((p \rightarrow(r \wedge q)) \rightarrow((p \rightarrow r) \vee(p \rightarrow q))\) (f) \(((p \rightarrow q) \rightarrow q) \rightarrow p\)

(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?

1\. Let \(X\) be the set of all students at a university. Let \(A\) be the set of students who are firstyear students, \(B\) the set of students who are second- year students, \(C\) the set of students who are in a discrete mathematics course, \(D\) the set of students who are international relations majors, \(E\) the set of students who went to a concert on Monday night, and \(F\) the set of students who studied until 2 AM on Tuesday. Express in set theoretic notation the following sets of students: (a) All second-year students in the discrete mathematics course. Sample Solution. \(\mid x \in X: x \in B\) and \(x \in C\\} .\) (b) All first-year students who studied until 2 AM on Tuesday. (c) All students who are international relations majors and went to the concert on Monday night. (d) All students who studied until 2 AM on Tuesday, are second-year students, and are not international relations majors. (e) All first- and second-year students who did not go to the concert on Monday night but are intemational relations majors. (f) All students who are first-year international relations majors or who studied until 2 AM on Tuesday. (g) All students who are first-or second-year students who went to a concert on Monday night. (h) All first-year students who are intemational relations majors or went to a concert on Monday night.

Let \(\phi=(p \rightarrow q) \rightarrow((r \wedge \neg s) \rightarrow q)\). For each of the following interpretations of \(p, q, r,\) and \(s,\) compute \(I(\phi)\) using the truth tables for \(\neg, v, \wedge, \rightarrow,\) and \(\leftrightarrow\) (a) \(I(p)=T, I(q)=T, I(r)=F,\) and \(I(s)=T\) (b) \(I(p)=T, I(q)=F, I(r)=T,\) and \(I(s)=F\) (c) \(I(p)=F, I(q)=T, I(r)=T,\) and \(I(s)=F\) (d) \(I(p)=F, I(q)=F, I(r)=T,\) and \(I(s)=F\)

The second stage of the procedure to "push negations inward" started with a formula whose only logical connectives are \(\neg, v,\) and \(\wedge\) and constructed a tautologically equivalent formula with negations applied only to proposition letters. (a) Write an algorithm describing exactly what is done. The algorithm should work on formulas as strings of symbols. To avoid what in this case is irrelevant detail, the program should assume that all proposition letters are one character long and that any symbol encountered, except for \((.), \wedge, v,\) and \(\neg,\) is a proposition letter. Assume that the formula contains no blanks. (It is perhaps easiest to consider the program as a function that is passed the original formula - a string-as a parameter, and then returns the equivalent formula with all the negations pushed inward. It is casiest to use recursion to handle many subformulas.) (b) Prove that your program from part (a) works. (Hint: if your program in part (a) uses recursion to handle subformulas, it is natural to do this proof by induction on formulas. However, the induction may not be straightforward.)

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