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Chapter 2: Problem 3

Show that \(\models \varphi(t) \leftrightarrow \forall x(x=t \rightarrow \varphi(x))\) if \(x \notin F V(t)\).

Short Answer

Expert verified
The formula is valid because \( x \notin FV(t) \), so \( \varphi(t) \) and \( \forall x(x=t \rightarrow \varphi(x)) \) are logically equivalent.

Step by step solution

01

Understand the Premise

We need to demonstrate that the formula \( \varphi(t) \leftrightarrow \forall x(x=t \rightarrow \varphi(x)) \) is valid given the condition that the variable \( x \) is not free in the term \( t \). This means that substituting \( x \) for any occurrence in \( t \) does not affect the truth of the formula.
02

Consider the Interpretation of \( \varphi(t) \)

The expression \( \varphi(t) \) has a specific interpretation which depends on how \( t \) is related to other variables and constants in the formula. In this premise, it implies that the formula holds considering the substitutions made only within the bounds and conditions that exclude the free variable \( x \) being part of \( t \).
03

Evaluate \( \forall x(x = t \rightarrow \varphi(x)) \)

This expression states that for every \( x \), if \( x \) is equal to \( t \), then \( \varphi(x) \) must be true. Because \( x \) does not appear in \( t \) (as it does not affect \( t \)), \( x = t \) only affects the scope of \( \varphi(x) \) directly.
04

Link the Two Expressions

Both expressions \( \varphi(t) \) and \( \forall x(x=t \rightarrow \varphi(x)) \) must logically evaluate to the same truth value. Given the condition \( x otin FV(t) \), \( t \) remains unaffected by changes to \( x \), meaning \( \varphi(t) \) itself reflects the value of \( \forall x(x=t \rightarrow \varphi(x)) \).
05

Conclude Logical Equivalence

With the condition \( x otin FV(t) \), changes to \( x \) do not affect \( t \), thus the equivalence \( \varphi(t) \leftrightarrow \forall x(x=t \rightarrow \varphi(x)) \) holds valid under all interpretations. This proves that the formula is universally valid under this condition.

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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 essential concept in logic. Two statements are logically equivalent if they always have the same truth value. This means that regardless of what the individual elements in the statements are, as long as the logical form is preserved, substituting different elements won't change the fact that both statements are true or false together.

In our problem, the equivalence \[\varphi(t) \leftrightarrow \forall x(x=t \rightarrow \varphi(x))\]
shows logical equivalence. Here, this means that the truth of \(\varphi(t)\) is equivalent to the truth of the statement \(\forall x(x=t \rightarrow \varphi(x))\).
  • Both expressions need to evaluate to the same truth value under the same conditions.
  • The condition \(x otin FV(t)\) ensures that changes in \(x\) do not alter \(t\), maintaining this equivalence.

Understanding logical equivalence requires recognizing that two different-looking expressions can fundamentally represent the same idea or truth, as demonstrated in this exercise.
Universal Quantification
Universal quantification is an element of logic that refers to the truth of a statement for all possible instances or elements of a particular set. When we say "for all \(x\)," we are universally quantifying over the set of possible values \(x\) can assume, and the statement must hold true for each one.

In this context, \(\forall x(x = t \rightarrow \varphi(x))\) means that if \(x\) is treated as somehow being equal to \(t\), then \(\varphi(x)\) should be true for all \(x\) that meet this condition. This kind of statement is incredibly powerful in logic because it asserts that under certain conditions, something must be true everywhere or for everything.
  • If any given instance \(x\) equals \(t\), then \(\varphi(x)\) must apply, rendering the original condition as universally valid as long as there's coherence in substitution.
  • The concept is aligned with logical equivalence, ensuring the statement is consistent across all potential substitutions of \(x\).

Universal quantification solidifies another aspect of logical foundations by affirming truths over vast sets of possibilities.
Free Variables
A free variable is a variable in a logical statement that is not bound by anything like a quantifier or specific substitution. It remains 'free' because it can take on any value, unlike bound variables, which are fixed in their roles within the context of the expression.

In logical expressions, distinguishing between free and bound variables is crucial for understanding the behavior of formulas under different conditions. In our exercise, the term \(t\) should not have the variable \(x\) free in it, denoted by \(x otin FV(t)\). This condition is crucial because:
  • The absence of \(x\) as a free variable in \(t\) ensures that any changes to \(x\) do not affect \(t\).
  • By ensuring \(x\) is not present in \(t\), we maintain the logical equivalence of \(\varphi(t)\) and \(\forall x(x = t \rightarrow \varphi(x))\). This consistency allows us to confidently assert that universal conditions hold.

Recognizing free variables helps to avoid unintended consequences that could arise from variable substitution, ensuring stability and validity within logical systems.

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Most popular questions from this chapter

Show that all propositional tautologies are true in all structures (of the right similarity type).

Show \(\models \forall x(\varphi \rightarrow \psi) \rightarrow(\forall x \varphi \rightarrow \forall x \psi) ; \models(\exists x \varphi \rightarrow \exists x \psi) \rightarrow \exists x(\varphi \rightarrow \psi)\) \(\models \forall x(\varphi \mapsto \psi) \rightarrow(\forall x \varphi \leftrightarrow \forall x \psi) ; \models(\forall x \varphi \rightarrow \exists x \psi) \leftrightarrow \exists x(\varphi \rightarrow \psi) ;\) \(\models(\exists x \varphi \rightarrow \forall x \psi) \rightarrow \forall x(\varphi \rightarrow \psi)\)

Consider \(\sigma_{1}=\forall x(x \sim x), \sigma_{2}=\forall x y(x \sim y \rightarrow y \sim x), \sigma_{3}=\forall x y z(x \sim\) \(y \wedge y \sim z \rightarrow x \sim z\) ). Show that if \(\mathfrak{Q} \models \sigma_{1} \wedge \sigma_{2} \wedge \sigma_{3}\), where \mathfrak{ } \(=\langle A, R)\), then \(R\) is an equivalence relation. N.B. \(x \sim y\) is a suggestive notation for the atom \(\bar{R}(x, y)\).

Find prenex normal forms for (a) \(\neg((\neg \forall x \varphi(x) \vee \forall x \psi(x)) \wedge(\exists x \sigma(x) \rightarrow \forall x \tau(x)),\), (b) \(\forall x \varphi(x) \leftrightarrow \exists x \psi(x)\), (c) \(\neg(\exists x \varphi(x, y) \wedge \forall y(\psi(y)) \rightarrow \varphi(x, x)) \rightarrow \exists x \forall y \sigma(x, y)))\), (d) \(((\forall x \varphi(x) \rightarrow \exists y \psi(x, y)) \rightarrow \psi(x, x)) \rightarrow \exists x \forall y \sigma(x, y)\).

Monadic predicate calculus has only unary predicate symbols (no identity). Consider \(\mathfrak{A}=\left\langle A, R_{1}, \ldots, R_{n}\right\rangle\) where all \(R_{i}\) are sets. Define \(a \sim\) \(b:=a \in R_{i} \Leftrightarrow b \in R_{i}\) for all \(i \leq n\). Show that \(\sim\) is an equivalence relation and that \(\sim\) has at most \(2^{n}\) equivalence classes. The equivalence class of \(a\) is denoted by \([a]\). Define \(B=A / \sim\) and \([a] \in S_{i} \Leftrightarrow a \in R_{i}, \mathfrak{B}=\) \(\left\langle B, S_{1}, \ldots, S_{n}\right\rangle .\) Show \(\mathfrak{A l} \models \sigma \Leftrightarrow \mathfrak{B} \models \sigma\) for all \(\sigma\) in the corresponding language. For such \(\sigma\) show \(\models \sigma \Leftrightarrow \mathfrak{A} \models \sigma\) for all \(\mathfrak{A}\) with at most \(2^{n}\) elements. Using this fact, outline a decision procedure for truth in monadic predicate calculus.
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