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Give an example of a universal set \(U\) and predicates \(P\) and \(Q\) such that \((\forall x P(x)) \rightarrow\) \((\forall x Q(x))\) is true but \(\forall x(P(x) \rightarrow Q(x))\) is false.

Short Answer

Expert verified
The universal set is \(U = \{1, 2, 3\}\), with predicates \(P(x)\) as "\(x\) is odd" and \(Q(x)\) as "\(x = 1\)."

Step by step solution

01

Define the Universal Set

Let's define a universal set \(U = \{1, 2, 3\}\). This set contains three elements that we'll use to evaluate predicates \(P\) and \(Q\).
02

Define Predicates P and Q

Define predicate \(P(x)\) as "\(x\) is an odd number." Thus, \(P(1)\) is true, \(P(2)\) is false, and \(P(3)\) is true. Now, define predicate \(Q(x)\) as "\(x = 1\)." Consequently, \(Q(1)\) is true, \(Q(2)\) is false, and \(Q(3)\) is false.
03

Evaluate \((\forall x P(x)) \rightarrow (\forall x Q(x))\)

\(\forall x P(x)\) is false because \(P(2)\) is false. In logical implication \(A \rightarrow B\), if \(A\) is false, \(A \rightarrow B\) is automatically true, regardless of the truth value of \(B\). Therefore, \((\forall x P(x)) \rightarrow (\forall x Q(x))\) is true.
04

Evaluate \(\forall x (P(x) \rightarrow Q(x))\)

We need to check each individual element in \(U\):\- \(P(1) \rightarrow Q(1)\) is true since both are true,\- \(P(2) \rightarrow Q(2)\) is true since \(P(2)\) is false,\- \(P(3) \rightarrow Q(3)\) is false since \(P(3)\) is true but \(Q(3)\) is false. \Hence, \(\forall x (P(x) \rightarrow Q(x))\) is false as one of the conditions, specifically \(x = 3\), fails.

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Key Concepts

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

Universal Set
In predicate logic, a universal set, denoted as \( U \), is the collection of all possible elements under consideration for a given discussion or problem. It's akin to a universe where every potential subject for your predicates resides.

For example, if we choose our universal set \( U = \{1, 2, 3\} \), it means that we will only be discussing numbers within this set. Each element in this universal set can be assessed using logical rules or predicates that we define. By clearly defining the universal set, we establish the boundary within which our logical statements and evaluations will take place.

By limiting our context to \( \{1, 2, 3\} \), we break down complex logical evaluations into manageable tasks by ensuring we're only considering these elements when testing predicates like \( P(x) \) or \( Q(x) \). This allows for precise and clear assessments of logical expressions. It's crucial to understand what elements our universal set includes to accurately interpret predicates and their logical implications.
Logical Implications
Logical implications are statements that express a condition of 'if...then...' between two logical assertions. In logical terms, an implication \( A \rightarrow B \) states "If \( A \) is true, then \( B \) must also be true."

This kind of statement has a fascinating rule: whenever \( A \) is false, \( A \rightarrow B \) is automatically true, regardless of whether \( B \) is true or false. This rule aligns with formal logic principles where a false premise can lead to any conclusion.

In our given exercise, we observe this in the implications \( (\forall x P(x)) \rightarrow (\forall x Q(x)) \). Since \( \forall x P(x) \) is false due to \( P(2) \) being false, the entire implication holds true. Understanding the mechanics of logical implications helps in reasoning about complex predicate logic problems, ensuring that each step follows logically from the previous conditions, thus maintaining integrity in logical reasoning.
Truth Values
In logic, each statement or predicate is assigned a truth value: either true or false. Truth values are pivotal in evaluating the validity of logical expressions and helping us reason through complex predicate logic.

For example, consider predicates \( P(x) \) and \( Q(x) \) with assigned values based on conditions: \( P(x) \) being true if \( x \) is odd and \( Q(x) \) true if \( x = 1 \). For each element in \( U \), we determine whether predicates hold true or false:
  • \( P(1) \) and \( Q(1) \) are both true.
  • \( P(2) \) is false and \( Q(2) \) is false.
  • \( P(3) \) is true, but \( Q(3) \) is false.
These truth values help us assess the entire logical statement \( \forall x (P(x) \rightarrow Q(x)) \), where each specific outcome affects whether the overall statement is true or false. For logical expressions to be fully understood, each part must be assessed for its truth value, as seen when \( P(3) \) failing to imply \( Q(3) \) renders the whole statement false.

Understanding truth values solidifies one’s grasp of logical arguments, making it easier to predict outcomes and validate logical expressions in predicate logic scenarios.

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Most popular questions from this chapter

Let \(U\) be the set of all problems on a comprehensive list of problems in science. Define four predicates over \(U\) by: \(P(x): x\) is a mathematics problem \(Q(x): x\) is difficult (according to some well-defined criterion: it does not matter for us What the criterion is) \(R(x): x\) is easy (according to some well-defined criterion) \(S(x): x\) is unsolvable (if you do not know what "unsolvable" means, do not worry about it here) Translate into English sentences each of the following formulas: (a) \(\forall x P(x)\) (b) \(\exists x Q(x)\) (c) \(\forall x(Q(x) \vee R(x))\) (d) \(\forall x(S(x) \rightarrow P(x))\) (e) \(\exists x(S(x) \wedge \neg P(x))\) (f) \(\neg(\forall x(\neg R(x) \vee S(x)))\) (g) \(\forall x(P(x) \rightarrow(Q(x) \leftrightarrow \neg R(x)))\) (h) \(\forall x \neg S(x)\) (i) \(\forall x(P(x) \rightarrow \neg S(x))\) (j) \(\forall x(P(x) \rightarrow(R(x) \vee S(x)))\) (k) \(\exists x(\neg Q(x) \wedge \neg R(x))\) (l) \(\exists x(R(x) \wedge S(x))\) (m) \( \forall x(Q(x) \leftrightarrow \neg R(x))\)

Given an array Values with \(n\) elements Values[0], Values[1],.... Values \([n-1 \mid\) each containing a real number, the following algorithm finds the sum of all the positive values in Values. Write an invariant for the loop. rollingSum \(=0\) for \(i=\phi, 2, \ldots, n-1\) if Values \([t]>0\) rolling Sum \(=\) rollingSum \(+\) Values \([i]\) Output rollingSum.

A restaurant displays the sign "Good food is not cheap." and a competing restaurant displays the sign "Cheap food is not good." Are the two restaurants saying the same thing?

Let \(A, B,\) and \(C\) be sets. (a) Prove that if \(A \subset B\) and \(B \subseteq C\), then \(A \subset C\). (b) Prove that if \(A \subseteq B\) and \(B \subset C,\) then \(A \subset C\). (c) Prove that if \(A \subseteq B\) and \(A \not \subseteq C,\) then \(B \nsubseteq C\).

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.)

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