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

Short Answer

Expert verified
The NNF for the negation is \(\forall x \exists y \exists z \neg P(x, y, z)\).

Step by step solution

01

Understand the Problem Statement

We need to find a formula in negation normal form (NNF) for the negation of the given formula \(\exists x \forall y \forall z (P(x, y, z))\). NNF requires that negations are applied directly to atomic propositions and only literals (\(P(x,y,z)\) or its negation) remain as arguments for disjunctions and conjunctions.
02

Negate the Given Formula

We start by applying negation to the entire formula: \(eg (\exists x \forall y \forall z (P(x, y, z)))\). To manipulate this negated expression, we use logical equivalence rules.
03

Use De Morgan's Laws and Quantifier Equivalence

Apply De Morgan's laws and quantifier equivalences: \(eg \exists x A(x) \equiv \forall x eg A(x)\) and \(eg \forall x A(x) \equiv \exists x eg A(x)\). Thus, for our expression, we proceed as follows:- \(eg (\exists x \forall y \forall z (P(x, y, z))) = \forall x eg (\forall y \forall z (P(x, y, z)))\) - Further apply: \(\forall x \exists y eg (\forall z (P(x, y, z)))\)- Finally: \(\forall x \exists y \exists z eg (P(x, y, z))\).
04

Confirm NNF Requirements

The negation normal form is achieved when negations only apply directly to propositions (and not to larger expressions of propositions). In the last expression \(\forall x \exists y \exists z eg (P(x, y, z))\), the negation applies directly to \(P(x, y, z)\), hence it is in negation normal form.

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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 foundational concept in logic, crucial for transforming formulas to their negation normal form (NNF). Two statements are logically equivalent if they always have the same truth value. This means that no matter the situation, if one statement is true, the other is also true, and vice versa. Understanding logical equivalence allows us to manipulate logical expressions without altering their inherent meaning.

In the context of our problem, we used logical equivalence to transform the formula \[eg (\exists x \forall y \forall z (P(x, y, z)))\] into \[\forall x \exists y \exists z eg (P(x, y, z))\]. We achieved this transformation through a series of logically equivalent steps, using specific rules like De Morgan's laws and quantifier equivalences. These transformations ensure that the essence of the original expression remains unchanged, while making it easier to understand or manipulate in its negation normal form. Logical equivalence is essential for simplifying complex logical statements and ensuring that transformations preserve the original meaning.
De Morgan's Laws
De Morgan's Laws are essential tools in logic for negating expressions, especially when dealing with conjunctions (\(\land\)) and disjunctions (\(\lor\)). Named after the mathematician Augustus De Morgan, these laws provide a way to distribute negation through a compound statement.

The laws state:
  • The negation of a conjunction is the disjunction of the negations: \(eg (A \land B) \equiv eg A \lor eg B\).
  • The negation of a disjunction is the conjunction of the negations: \(eg (A \lor B) \equiv eg A \land eg B\).
In the given exercise, De Morgan's laws facilitated the distribution of negation across a nested logical structure. When we negated the expression \(\exists x \forall y \forall z (P(x, y, z))\), De Morgan's laws helped us to deconstruct it and apply negation effectively. Understanding and applying these laws allows for the simplification of complex logical formulas, ensuring that they conform to the rules of negation normal form.
Quantifier Equivalence
Quantifier equivalence involves understanding how existential (\(\exists\)) and universal (\(\forall\)) quantifiers interact with negation. This knowledge is pivotal when deducing the negation normal form of logical expressions involving quantifiers.

The rules for quantifier equivalence state:
  • \(eg \exists x A(x) \equiv \forall x eg A(x)\)
  • \(eg \forall x A(x) \equiv \exists x eg A(x)\)
These rules show that negating a statement with an existential quantifier converts it into a universal quantifier, and vice versa. This interchange is necessary for fitting expressions into a negation normal form. In our solution, we originally negated \(\exists x \forall y \forall z (P(x, y, z))\). By understanding quantifier equivalence, we converted it step by step, preserving its logical truth while pushing the negation inward to a position where it directly applies to the proposition \(P(x, y, z)\).

Quantifier equivalence is crucial for logical transformations that maintain the meaning of a statement while altering its structure, especially when simplifying expressions for specific logical forms like NNF.

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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 \(I(s)=T\)

For any two integers \(m\) and \(n,\) we say \(m\) divides \(n\) if there is an integer \(k\) such that \(n=\) \(m k\). (Many programming languages give easy ways to say that, such as \(n \% m=0\) or \(n\) div \(m=0 .\) ) Define \(D i v(m, n)\) to be \(m\) divides \(n\). Translate each of the following propositions and quantified formulas into a clear English sentence. Label each as being true or false, with the universe as the set \(\mathbb{Z}\). (a) \(\operatorname{Div}(5,7)\) (b) \(\operatorname{Div}(4,16)\) (c) \(\operatorname{Div}(16,4)\) (d) \(\operatorname{Div}(-8,0)\) (e) \(\forall m(\forall n(\operatorname{Div}(m, n)))\) (f) \(\forall n(\operatorname{Div}(1, n))\) (g) \(\forall m(\operatorname{Div}(m, 0))\) (h) \(\forall m(\forall n(\operatorname{Div}(m, n) \rightarrow \operatorname{Div}(n, m)))\) (i) \(\forall m(\forall n(\forall p((D i v(m, n) \wedge \operatorname{Div}(n, p)) \rightarrow \operatorname{Div}(m, p))))\) (j) \(\forall m(\forall n((D i v(m, n) \wedge \operatorname{Div}(n, m)) \rightarrow m=n))\)

This problem concerns the following six sets: $$\begin{array}{c}A=\\{0,2,4,6\\} \quad B=(1,3,5) \quad C=\\{0,1,2,3,4,5,6,7\\} \\\D=\emptyset \quad E=\mathbb{N} \quad F=\\{10,2,4,6\\} \mid\end{array}$$ (a) What sets are subsets of \(A\) ? (b) What sets are subsets of \(B\) ? (c) What sets are subsets of \(C\) ? (d) What sets are subsets of \(D\) ? (c) What sets are subsets of \(E\) ? (f) What sets are subsets of \(F\) ?

(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 \(n\).

Find the expression tree for the formula $$ \neg(p \wedge q) \leftrightarrow(\neg p \vee \neg q) $$ Evaluate the expression tree for all possible pairs of truth values for \(p\) and \(q\). Use these evaluations to prove this formula is a tautology.

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