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

Short Answer

Expert verified
The negation normal form is \( \exists x \forall y ((P(x, y) \wedge Q(x, y)) \land \neg R(x, y)) \).

Step by step solution

01

Apply Negation to the Entire Statement

Start by applying negation to the entire formula. We are given \( \forall x \exists y((P(x, y) \wedge Q(x, y)) \rightarrow R(x, y)) \) and need to find its negation: \( eg \forall x \exists y((P(x, y) \wedge Q(x, y)) \rightarrow R(x, y)) \).
02

Use Negation of Quantifiers

Use the law of negation for quantifiers. The negation of \( \forall x \exists y \) is \( \exists x eg \exists y \). Thus, the statement becomes \( \exists x \forall y eg ((P(x, y) \wedge Q(x, y)) \rightarrow R(x, y)) \).
03

Replace the Implication with Disjunction

Use the logical equivalence \( (A \rightarrow B) \equiv (eg A \lor B) \) to replace the implication inside the negation. Our expression is now \( \exists x \forall y eg (eg (P(x, y) \wedge Q(x, y)) \lor R(x, y)) \).
04

Distribute Negation Inside Parentheses

Apply De Morgan's laws to move negation inwards. The expression becomes \( \exists x \forall y ((P(x, y) \wedge Q(x, y)) \land eg R(x, y)) \) because negating \( eg A \lor B \) gives \( A \land eg B \).
05

Simplify to Negation Normal Form

Now the formula is in negation normal form because all negations are directly applied to atomic propositions and the formula uses only the logical connectives \( \land \) and \( \lor \). The final negation normal form is \( \exists x \forall y ((P(x, y) \wedge Q(x, y)) \land eg R(x, y)) \).

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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 logic, seen when two statements always have the same truth value. In simpler terms, it means that whenever one statement is true, the other must be true, and vice versa. Logical equivalence is a powerful tool because it allows us to transform complex expressions into simpler or more useful forms, which is often essential in mathematical proofs and computer science. This process involves substituting segments of a logical formula with equivalent segments until a desired form is achieved. In this particular exercise, logical equivalence is used to replace the implication within the formula with a disjunction of the form \((eg A \lor B)\), facilitating further transformations using rules like De Morgan's Laws.
Quantifiers
Quantifiers are symbols used in logic to express statements involving 'all' or 'some', such as the phrases 'for all' and 'there exists'. The two most common quantifiers are the universal quantifier \(\forall\) and the existential quantifier \(\exists\).

- \(\forall x\) (for all x): This quantifies that a statement holds true for every possible value of x.
- \(\exists x\) (there exists x): This indicates there is at least one value of x for which the statement holds true.

In the solved exercise, negation impacts these quantifiers. Using the rule of negation, we swap \(\forall\) and \(\exists\): \(eg \forall x\) becomes \(\exists x eg\), and \(eg \exists x\) becomes \(\forall x eg\). Adjusting the quantifiers could change the meaning of statements significantly, emphasizing the importance of accurate transformations in logical equivalence.
De Morgan's Laws
De Morgan's Laws are crucial for handling negations within logical statements and are key to simplifying complex logical expressions. These laws provide rules for distributing negations across conjunctions (\(\land\)) and disjunctions (\(\lor\)). They state:

- The negation of a conjunction \((eg(A \land B))\) is equivalent to the disjunction of the negations \((eg A \lor eg B)\).
- The negation of a disjunction \((eg(A \lor B))\) is equivalent to the conjunction of the negations \((eg A \land eg B)\).

In the original exercise, De Morgan's Laws are applied to a negated disjunction resulting from the logical equivalence transform. The law is used to remove negations that cover whole expressions, instead pushing them further into the formula, thus bringing the formula closer to the standard negation normal form.
Negation
Negation involves the reversal of the truth value of a proposition or expression, transforming true into false and vice versa. Understanding how negation operates is fundamental to manipulating logical expressions, especially when working toward negation normal form.

Negation is first applied in the exercise to the entire logical structure, serving as the initial step in altering the formula. This task necessitates further manipulation due to the involvement of implications and quantifiers. Each negation needs careful handling due to its powerful role in altering meaning. The interaction of negation with other logical operations, as modeled by De Morgan's Laws and the transformation of quantifiers, ensures that the final expression maintains logical equivalence to the original statement—just with the negation correctly influencing the whole structural layout.

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Most popular questions from this chapter

Find formulas in CNF equivalent to each of the following formulas: (a) \(\neg(p \wedge T)\) (b) \(((p \rightarrow q) \rightarrow r) \rightarrow F\) (c) \(((p \rightarrow q) \rightarrow r) \rightarrow T\) (d) \((p \leftrightarrow q) \leftrightarrow r\) (e) \(\neg(p \leftrightarrow q) \leftrightarrow r\) (f) \(((p \vee q) \rightarrow r) \wedge(r \rightarrow \neg(p \vee q))\) (g) \((\neg r) \rightarrow(((p \vee q) \rightarrow r) \rightarrow \neg q)\)

Find the expression tree for the formula $$ ((\neg(p \wedge q)) \vee(\neg(q \wedge r))) \wedge((\neg(p \leftrightarrow(\neg(\neg s)))) \vee((r \wedge s) \vee(\neg q))) . $$ Evaluate the expression tree if proposition \(p\) is \(F\), proposition \(q\) is \(T\), proposition \(r\) is \(F\), and proposition \(s\) is \(T\).

Write three descriptions of the elements of the set 12,5,8,11,14\(\\}\)

Translate each of the following quantified formulas into an English sentence where the universal set is \(\mathbb{R}\). Label each as true or false. (a) \(\forall x(\exists y(x y=x))\) (b) \(\forall y(\exists x(x y=x))\) (c) \(\forall x(\exists y(x y=1))\) (d) \(\exists y(\forall x \neq 0(x y=1))\) (e) \(\exists x(\forall y(x y=x))\) (f) \((\forall x(x \neq 0 \rightarrow \exists y(x y=1))\)

(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 \(n\).

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