First-Order Logic Exercises

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Your Name

Published

February 5, 2025

Introduction

This document covers exercises on first-order logic, focusing on syntax and semantics. We will explore how to define universes, signatures, and interpretations for various properties, and then translate these properties into first-order formulas. The main objectives are to understand how to represent objects and their relationships in first-order logic, and to learn how to express complex properties using quantification and logical connectives. Key concepts include the representation of graphs, prime numbers, trees, and finite strings. We will also touch upon the limitations of first-order logic in expressing certain properties, such as the existence of infinitely many objects without an order relation and the transitive closure of a relation.

Graphs and Cliques

Problem Statement

Define a first-order formula to represent the property "This graph is a clique."

Solution

Universe: The set of nodes in the graph.
Signature: A single binary relation symbol $E$, representing edges.
Interpretation: $E(x, y)$ is true if there is an edge between nodes $x$ and $y$.

A graph is a clique if every pair of distinct nodes is connected by an edge.

Formula:

This formula states that for every pair of nodes $x$ and $y$, either there is an edge between them or they are the same node. \[\phi \equiv \forall x \forall y (E(x, y) \lor x = y)\]

This formula states that for every pair of nodes $x$ and $y$, either there is an edge between them or they are the same node.

Alternative Formula (excluding self-loops):

This formula states that for every pair of distinct nodes $x$ and $y$, there is an edge between them. \[\phi \equiv \forall x \forall y (x \neq y \rightarrow E(x, y))\]

This formula states that for every pair of distinct nodes $x$ and $y$, there is an edge between them.

Prime Numbers

Problem Statement

Define a first-order formula to represent the property "There exist infinitely many prime numbers."

Solution

Universe: The set of natural numbers, $\mathbb{N}$.
Signature:
- A ternary relation symbol $R_\cdot$ for multiplication, where $R_\cdot(x,y,z)$ represents $x = y \cdot z$.
- A binary relation symbol $O$ for the natural ordering, where $O(x,y)$ represents $x < y$.
- (Alternative) A unary relation symbol $P$ for prime numbers, where $P(x)$ represents that $x$ is a prime number.
Interpretation:
- $R_\cdot(x, y, z)$ is true if $x = y \cdot z$.
- $O(x, y)$ is true if $x < y$.
- (Alternative) $P(x)$ is true if $x$ is a prime number.

A number $x$ is prime if for all $y$ and $z$, if $x = y \cdot z$, then either $y = 1$ or $z = 1$.

Defining the number 1:

This formula defines the number 1 as the unique element $t$ such that for all $v$, $t \cdot v = v$. \[\text{One}(t) \equiv \forall v (R_\cdot(t, v, v))\]

This formula defines the number 1 as the unique element $t$ such that for all $v$, $t \cdot v = v$.

Formula for prime numbers (using multiplication):

This formula defines a prime number $x$ using the multiplication relation. \[\alpha(x) \equiv \forall y \forall z (R_\cdot(x, y, z) \rightarrow (\exists t (\text{One}(t) \land (y = t \lor z = t))))\]

Alternative formula for prime numbers (using a unary relation):

This formula defines a prime number $x$ directly using a unary relation $P$. \[\alpha(x) \equiv P(x)\]

Formula for infinitely many primes:

This formula states that for every number $x$, there exists a number $y$ greater than $x$ such that $y$ is prime. \[\phi \equiv \forall x \exists y (O(x, y) \land \alpha(y))\]

This formula states that for every number $x$, there exists a number $y$ greater than $x$ such that $y$ is prime.

The formula for infinitely many primes relies on the ability to express order. Without an order relation, it is impossible to state the existence of infinitely many objects in first-order logic. This is because "infinitely many" implies a statement about cardinality, which requires quantification over sets, a feature not available in first-order logic.

Least Common Ancestor in a Tree

Problem Statement

Define a first-order formula to represent the property "$Z$ is the least common ancestor of $X$ and $Y$ in this tree."

Solution

Universe: The set of nodes in the tree.
Signature:
- A binary relation symbol $A$ for the ancestor relationship.
- (Alternative) A binary relation symbol $E$ for the parent-child relationship.
Interpretation:
- $A(x, y)$ is true if $x$ is an ancestor of $y$.
- (Alternative) $E(x, y)$ is true if $x$ is the parent of $y$.

Formula (using ancestor relation):

This formula defines the least common ancestor $Z$ of $X$ and $Y$ using the ancestor relation $A$. \[\phi(X, Y, Z) \equiv A(Z, X) \land A(Z, Y) \land \forall Z' ((A(Z', X) \land A(Z', Y)) \rightarrow A(Z', Z))\]

This formula states that $Z$ is an ancestor of both $X$ and $Y$, and for any other common ancestor $Z'$ of $X$ and $Y$, $Z'$ is an ancestor of $Z$.

If we use the parent-child relation $E$ instead of the ancestor relation $A$, it is not possible to define the least common ancestor in first-order logic without knowing the height of the tree. The ancestor relation is the transitive closure of the parent-child relation, and transitive closures cannot generally be expressed in first-order logic. This is because expressing transitive closure would require an unbounded number of conjunctions or disjunctions to represent paths of arbitrary length in the tree, which is not allowed in first-order logic.

Finite Strings

Problem Statement

Define a first-order formula to represent the property "In this finite string, every A is immediately followed by a B."

Solution

Universe: The set of positions in the string, $\{1, 2, \dots, n\}$, where $n$ is the length of the string.
Signature:
- Unary relation symbols $A$, $B$, and $C$ representing positions labeled with ‘A’, ‘B’, and ‘C’, respectively.
- A binary relation symbol $S$ for the successor relation (or a binary relation symbol $O$ for order).
Interpretation:
- $A(x)$ is true if position $x$ is labeled with ‘A’.
- $B(x)$ is true if position $x$ is labeled with ‘B’.
- $C(x)$ is true if position $x$ is labeled with ‘C’.
- $S(x, y)$ is true if $y = x + 1$.
- (Alternative) $O(x,y)$ is true if $x < y$.

In this example, the string is "ABBABC". Positions 1, 4, and 7 are labeled with ‘A’, positions 2, 3, and 5 are labeled with ‘B’, and position 6 is labeled with ‘C’.

Formula:

This formula states that for every position $x$, if $x$ is labeled with ‘A’, then there exists a position $y$ such that $y$ is the successor of $x$ and $y$ is labeled with ‘B’. \[\phi \equiv \forall x (A(x) \rightarrow \exists y (S(x, y) \land B(y)))\]

This formula states that for every position $x$, if $x$ is labeled with ‘A’, then there exists a position $y$ such that $y$ is the successor of $x$ and $y$ is labeled with ‘B’.

Alternative Formula (using order):

This formula states that for every position $x$, if $x$ is labeled with ‘A’, then there exists a position $y$ such that $y$ is immediately after $x$ and labeled with ‘B’. \[\phi \equiv \forall x (A(x) \rightarrow \exists y (O(x,y) \land B(y) \land \forall z (O(x,z) \land O(z,y) \rightarrow \text{false})))\]

This formula states that for every position $x$, if $x$ is labeled with ‘A’, then there exists a position $y$ such that $y$ is greater than $x$, $y$ is labeled with ‘B’, and there is no position $z$ between $x$ and $y$.

Finite strings can be represented using unary relations for each letter, partitioning the set of positions. This approach is useful for modeling computations of systems where positions represent time points and letters represent events or actions. For example, in a coffee machine, the sequence of events (inserting coins, selecting a drink, dispensing the drink) can be represented as a string, where each position in the string corresponds to a time point and each letter corresponds to an event. The unary relations can represent the occurrence of each event at each time point.

Conclusion

These exercises demonstrate how to represent various properties in first-order logic by defining appropriate universes, signatures, and interpretations. We have seen that the choice of signature can significantly affect the expressiveness of the logic. For example, the ability to express order is crucial for stating the existence of infinitely many objects, and the ancestor relation (a transitive closure) cannot generally be defined in terms of the parent-child relation in first-order logic. We also observed that representing functions over finite domains can be efficiently done using unary relations, as in the case of finite strings.

Follow-up Questions:

How would you represent the property "This graph is connected" in first-order logic?
- Answer: This property cannot be expressed in first-order logic. Connectivity requires the ability to express the transitive closure of the edge relation, which, as discussed, is not possible in first-order logic.
Can you define a first-order formula for the property "This number is a power of 2"?
- Answer: This is not directly expressible in first-order logic without additional predicates or functions. One might attempt to define it using repeated multiplication by 2, but this would require an unbounded number of quantifiers, which is not allowed in first-order logic.
How would you modify the string representation if the alphabet had infinitely many letters?
- Answer: If the alphabet were infinite, we could not use a finite set of unary relations to represent each letter. We would need a different approach, such as introducing a binary relation $L(x, y)$ where $x$ is a position and $y$ is a letter from the infinite alphabet. However, this would require quantifying over an infinite domain of letters, which changes the nature of the logic.
What are the limitations of first-order logic in expressing properties of finite strings?
- Answer: First-order logic can express properties related to the order and labeling of positions in a finite string. However, it cannot express properties that require counting beyond a fixed bound (e.g., "more A’s than B’s") or properties that involve recursion or unbounded iteration (e.g., "the string is a palindrome").
(Detail missing: teacher explanation unclear) How to represent a property that requires counting in first-order logic? Suggestion: Introduce a relation for counting or a way to represent numbers.
- Answer: First-order logic cannot express properties that require counting beyond a fixed bound. To represent counting, one would need to introduce additional structure, such as arithmetic operations or a way to represent numbers and their relationships. However, this would lead to a more complex logic than standard first-order logic.
(Detail missing: teacher explanation unclear) Can we define a formula for the transitive closure of a relation in first-order logic? Suggestion: No, transitive closure cannot be generally defined in first-order logic.
- Answer: No, the transitive closure of a relation cannot be generally defined in first-order logic. This limitation is a fundamental aspect of first-order logic and is related to its inability to express recursion or unbounded iteration.

--- title: "First-Order Logic Exercises" author: "Your Name" date: "2025-02-05" format: html: toc: true # Table of Contents toc-depth: 2 code-tools: true theme: cosmo # Or "journal" for Distill-like minimalism --- # Introduction This document covers exercises on first-order logic, focusing on syntax and semantics. We will explore how to define universes, signatures, and interpretations for various properties, and then translate these properties into first-order formulas. The main objectives are to understand how to represent objects and their relationships in first-order logic, and to learn how to express complex properties using quantification and logical connectives. Key concepts include the representation of graphs, prime numbers, trees, and finite strings. We will also touch upon the limitations of first-order logic in expressing certain properties, such as the existence of infinitely many objects without an order relation and the transitive closure of a relation. # Graphs and Cliques ## Problem Statement Define a first-order formula to represent the property \"This graph is a clique.\" ## Solution - **Universe**: The set of nodes in the graph. - **Signature**: A single binary relation symbol $E$, representing edges. - **Interpretation**: $E(x, y)$ is true if there is an edge between nodes $x$ and $y$. ::: tcolorbox A graph is a clique if every pair of distinct nodes is connected by an edge. ::: :::: tcolorbox ::: center ::: :::: **Formula**: ::: tcolorbox This formula states that for every pair of nodes $x$ and $y$, either there is an edge between them or they are the same node. $$\phi \equiv \forall x \forall y (E(x, y) \lor x = y)$$ ::: This formula states that for every pair of nodes $x$ and $y$, either there is an edge between them or they are the same node. **Alternative Formula (excluding self-loops)**: ::: tcolorbox This formula states that for every pair of distinct nodes $x$ and $y$, there is an edge between them. $$\phi \equiv \forall x \forall y (x \neq y \rightarrow E(x, y))$$ ::: This formula states that for every pair of distinct nodes $x$ and $y$, there is an edge between them. # Prime Numbers ## Problem Statement Define a first-order formula to represent the property \"There exist infinitely many prime numbers.\" ## Solution - **Universe**: The set of natural numbers, $\mathbb{N}$. - **Signature**: - A ternary relation symbol $R_\cdot$ for multiplication, where $R_\cdot(x,y,z)$ represents $x = y \cdot z$. - A binary relation symbol $O$ for the natural ordering, where $O(x,y)$ represents $x < y$. - (Alternative) A unary relation symbol $P$ for prime numbers, where $P(x)$ represents that $x$ is a prime number. - **Interpretation**: - $R_\cdot(x, y, z)$ is true if $x = y \cdot z$. - $O(x, y)$ is true if $x < y$. - (Alternative) $P(x)$ is true if $x$ is a prime number. ::: tcolorbox A number $x$ is prime if for all $y$ and $z$, if $x = y \cdot z$, then either $y = 1$ or $z = 1$. ::: **Defining the number 1**: ::: tcolorbox This formula defines the number 1 as the unique element $t$ such that for all $v$, $t \cdot v = v$. $$\text{One}(t) \equiv \forall v (R_\cdot(t, v, v))$$ ::: This formula defines the number 1 as the unique element $t$ such that for all $v$, $t \cdot v = v$. **Formula for prime numbers (using multiplication)**: ::: tcolorbox This formula defines a prime number $x$ using the multiplication relation. $$\alpha(x) \equiv \forall y \forall z (R_\cdot(x, y, z) \rightarrow (\exists t (\text{One}(t) \land (y = t \lor z = t))))$$ ::: **Alternative formula for prime numbers (using a unary relation)**: ::: tcolorbox This formula defines a prime number $x$ directly using a unary relation $P$. $$\alpha(x) \equiv P(x)$$ ::: **Formula for infinitely many primes**: ::: tcolorbox This formula states that for every number $x$, there exists a number $y$ greater than $x$ such that $y$ is prime. $$\phi \equiv \forall x \exists y (O(x, y) \land \alpha(y))$$ ::: This formula states that for every number $x$, there exists a number $y$ greater than $x$ such that $y$ is prime. ::: tcolorbox The formula for infinitely many primes relies on the ability to express order. Without an order relation, it is impossible to state the existence of infinitely many objects in first-order logic. This is because \"infinitely many\" implies a statement about cardinality, which requires quantification over sets, a feature not available in first-order logic. ::: # Least Common Ancestor in a Tree ## Problem Statement Define a first-order formula to represent the property \"$Z$ is the least common ancestor of $X$ and $Y$ in this tree.\" ## Solution - **Universe**: The set of nodes in the tree. - **Signature**: - A binary relation symbol $A$ for the ancestor relationship. - (Alternative) A binary relation symbol $E$ for the parent-child relationship. - **Interpretation**: - $A(x, y)$ is true if $x$ is an ancestor of $y$. - (Alternative) $E(x, y)$ is true if $x$ is the parent of $y$. :::: tcolorbox ::: center ::: :::: **Formula (using ancestor relation)**: ::: tcolorbox This formula defines the least common ancestor $Z$ of $X$ and $Y$ using the ancestor relation $A$. $$\phi(X, Y, Z) \equiv A(Z, X) \land A(Z, Y) \land \forall Z' ((A(Z', X) \land A(Z', Y)) \rightarrow A(Z', Z))$$ ::: This formula states that $Z$ is an ancestor of both $X$ and $Y$, and for any other common ancestor $Z'$ of $X$ and $Y$, $Z'$ is an ancestor of $Z$. ::: tcolorbox If we use the parent-child relation $E$ instead of the ancestor relation $A$, it is not possible to define the least common ancestor in first-order logic without knowing the height of the tree. The ancestor relation is the transitive closure of the parent-child relation, and transitive closures cannot generally be expressed in first-order logic. This is because expressing transitive closure would require an unbounded number of conjunctions or disjunctions to represent paths of arbitrary length in the tree, which is not allowed in first-order logic. ::: # Finite Strings ## Problem Statement Define a first-order formula to represent the property \"In this finite string, every A is immediately followed by a B.\" ## Solution - **Universe**: The set of positions in the string, $\{1, 2, \dots, n\}$, where $n$ is the length of the string. - **Signature**: - Unary relation symbols $A$, $B$, and $C$ representing positions labeled with 'A', 'B', and 'C', respectively. - A binary relation symbol $S$ for the successor relation (or a binary relation symbol $O$ for order). - **Interpretation**: - $A(x)$ is true if position $x$ is labeled with 'A'. - $B(x)$ is true if position $x$ is labeled with 'B'. - $C(x)$ is true if position $x$ is labeled with 'C'. - $S(x, y)$ is true if $y = x + 1$. - (Alternative) $O(x,y)$ is true if $x < y$. :::: tcolorbox ::: center ::: In this example, the string is \"ABBABC\". Positions 1, 4, and 7 are labeled with 'A', positions 2, 3, and 5 are labeled with 'B', and position 6 is labeled with 'C'. :::: **Formula**: ::: tcolorbox This formula states that for every position $x$, if $x$ is labeled with 'A', then there exists a position $y$ such that $y$ is the successor of $x$ and $y$ is labeled with 'B'. $$\phi \equiv \forall x (A(x) \rightarrow \exists y (S(x, y) \land B(y)))$$ ::: This formula states that for every position $x$, if $x$ is labeled with 'A', then there exists a position $y$ such that $y$ is the successor of $x$ and $y$ is labeled with 'B'. **Alternative Formula (using order)**: ::: tcolorbox This formula states that for every position $x$, if $x$ is labeled with 'A', then there exists a position $y$ such that $y$ is immediately after $x$ and labeled with 'B'. $$\phi \equiv \forall x (A(x) \rightarrow \exists y (O(x,y) \land B(y) \land \forall z (O(x,z) \land O(z,y) \rightarrow \text{false})))$$ ::: This formula states that for every position $x$, if $x$ is labeled with 'A', then there exists a position $y$ such that $y$ is greater than $x$, $y$ is labeled with 'B', and there is no position $z$ between $x$ and $y$. ::: tcolorbox Finite strings can be represented using unary relations for each letter, partitioning the set of positions. This approach is useful for modeling computations of systems where positions represent time points and letters represent events or actions. For example, in a coffee machine, the sequence of events (inserting coins, selecting a drink, dispensing the drink) can be represented as a string, where each position in the string corresponds to a time point and each letter corresponds to an event. The unary relations can represent the occurrence of each event at each time point. ::: # Conclusion These exercises demonstrate how to represent various properties in first-order logic by defining appropriate universes, signatures, and interpretations. We have seen that the choice of signature can significantly affect the expressiveness of the logic. For example, the ability to express order is crucial for stating the existence of infinitely many objects, and the ancestor relation (a transitive closure) cannot generally be defined in terms of the parent-child relation in first-order logic. We also observed that representing functions over finite domains can be efficiently done using unary relations, as in the case of finite strings. **Follow-up Questions**: 1. How would you represent the property \"This graph is connected\" in first-order logic? - *Answer:* This property cannot be expressed in first-order logic. Connectivity requires the ability to express the transitive closure of the edge relation, which, as discussed, is not possible in first-order logic. 2. Can you define a first-order formula for the property \"This number is a power of 2\"? - *Answer:* This is not directly expressible in first-order logic without additional predicates or functions. One might attempt to define it using repeated multiplication by 2, but this would require an unbounded number of quantifiers, which is not allowed in first-order logic. 3. How would you modify the string representation if the alphabet had infinitely many letters? - *Answer:* If the alphabet were infinite, we could not use a finite set of unary relations to represent each letter. We would need a different approach, such as introducing a binary relation $L(x, y)$ where $x$ is a position and $y$ is a letter from the infinite alphabet. However, this would require quantifying over an infinite domain of letters, which changes the nature of the logic. 4. What are the limitations of first-order logic in expressing properties of finite strings? - *Answer:* First-order logic can express properties related to the order and labeling of positions in a finite string. However, it cannot express properties that require counting beyond a fixed bound (e.g., \"more A's than B's\") or properties that involve recursion or unbounded iteration (e.g., \"the string is a palindrome\"). 5. (Detail missing: teacher explanation unclear) How to represent a property that requires counting in first-order logic? Suggestion: Introduce a relation for counting or a way to represent numbers. - *Answer:* First-order logic cannot express properties that require counting beyond a fixed bound. To represent counting, one would need to introduce additional structure, such as arithmetic operations or a way to represent numbers and their relationships. However, this would lead to a more complex logic than standard first-order logic. 6. (Detail missing: teacher explanation unclear) Can we define a formula for the transitive closure of a relation in first-order logic? Suggestion: No, transitive closure cannot be generally defined in first-order logic. - *Answer:* No, the transitive closure of a relation cannot be generally defined in first-order logic. This limitation is a fundamental aspect of first-order logic and is related to its inability to express recursion or unbounded iteration.