Introduction to Logic
Logic is the foundation of mathematical reasoning and philosophy. It is the study of principles of valid inference and demonstration. Logic allows us to construct and analyze arguments, test their validity, and deduce conclusions from premises. There are two main types of logic: propositional logic and predicate logic.
Importance of Logic in Mathematics and Philosophy
Logic plays a critical role in both mathematics and philosophy. In mathematics, logic helps in proving theorems, analyzing algorithms, and formalizing concepts. In philosophy, logic is used to clarify reasoning, construct arguments, and explore metaphysical and ethical questions.
Propositional Logic
Propositional logic (also known as sentential logic) deals with propositions and their relationships through logical connectives. A proposition is a statement that is either true or false, but not both.
Example of propositions:
- “It is raining.”
- “2 + 2 = 4.”
Propositions and Truth Values
A proposition is any declarative statement that is either true or false. Each proposition has a truth value:
- True ($T$)
- False ($F$)
For example, “2 is greater than 1” is a proposition with a truth value of true ($T$), while “5 is less than 3” has a truth value of false ($F$).
Logical Connectives (AND, OR, NOT, IF-THEN, IFF)
Logical connectives are used to combine propositions to form more complex statements.
- AND ($\land$): The statement $p \land q$ is true if both $p$ and $q$ are true.
- OR ($\lor$): The statement $p \lor q$ is true if at least one of $p$ or $q$ is true.
- NOT ($\neg$): The statement $\neg p$ is true if $p$ is false.
- IF-THEN ($\rightarrow$): The statement $p \rightarrow q$ (implication) is false only when $p$ is true and $q$ is false.
- IFF ($\leftrightarrow$): The statement $p \leftrightarrow q$ (if and only if) is true if $p$ and $q$ have the same truth value.
Truth Tables
Truth tables show the truth value of a compound proposition for every possible combination of truth values of its components.
Example:
Answer:
Step 1: Given Data:
Let $p$ be “It is raining” and $q$ be “It is cloudy.”
Step 2: Solution:
| $p$ | $q$ | $p \land q$ | $p \lor q$ | $\neg p$ |
|---|---|---|---|---|
| $T$ | $T$ | $T$ | $T$ | $F$ |
| $T$ | $F$ | $F$ | $T$ | $F$ |
| $F$ | $T$ | $F$ | $T$ | $T$ |
| $F$ | $F$ | $F$ | $F$ | $T$ |
Step 3: Final Answer:
The truth table shows the possible truth values for $p \land q$, $p \lor q$, and $\neg p$.
Tautologies and Contradictions
A tautology is a proposition that is always true, regardless of the truth values of its components.
A contradiction is a proposition that is always false.
Example:
- Tautology: $p \lor \neg p$
- Contradiction: $p \land \neg p$
Logical Equivalence
Two propositions are logically equivalent if they have the same truth value in every possible situation. We denote logical equivalence as $p \equiv q$.
Example:
$p \rightarrow q \equiv \neg p \lor q$
De Morgan’s Laws in Propositional Logic
De Morgan’s Laws describe how negation interacts with conjunction and disjunction:
- $ \neg (p \land q) \equiv \neg p \lor \neg q $
- $ \neg (p \lor q) \equiv \neg p \land \neg q $
Implication and Biconditional Statements
Implication ($p \rightarrow q$) states that if $p$ is true, then $q$ must also be true.
Biconditional ($p \leftrightarrow q$) means that $p$ is true if and only if $q$ is true.
Propositional Satisfiability
A proposition is satisfiable if there is some assignment of truth values to its components that makes the proposition true. Otherwise, it is unsatisfiable.
Predicate Logic (First-Order Logic)
Predicate logic extends propositional logic by dealing with predicates and quantifiers. A predicate is a statement that contains variables and becomes a proposition when specific values are substituted for these variables.
Example:
Predicate: “x is greater than 5”
With $x = 7$, the predicate becomes the proposition “7 is greater than 5” (true).
Quantifiers (Universal and Existential)
Universal quantifier ($\forall$) states that a predicate is true for all elements in a given domain.
Existential quantifier ($\exists$) states that there is at least one element in the domain for which the predicate is true.
Example:
- Universal quantifier: $\forall x (x > 0)$ means “for all $x$, $x$ is greater than 0.”
- Existential quantifier: $\exists x (x > 0)$ means “there exists an $x$ such that $x$ is greater than 0.”
Predicate Symbols and Functions
In predicate logic, predicates are often denoted as $P(x)$, where $x$ is the variable. A function in predicate logic assigns values to these variables.
Interpretation of Predicates
An interpretation in predicate logic assigns meanings to the symbols used. This includes assigning truth values to predicates based on a particular domain.
Rules of Inference in Propositional Logic
Rules of inference allow us to derive conclusions from premises. Common rules include:
- Modus Ponens: $p \rightarrow q$, $p$, therefore $q$.
- Modus Tollens: $p \rightarrow q$, $\neg q$, therefore $\neg p$.
Rules of Inference in Predicate Logic
Rules of inference in predicate logic are similar to those in propositional logic but involve quantifiers. For example, the universal instantiation rule allows us to infer $P(x)$ from $\forall x P(x)$.
Logical Arguments and Validity
A logical argument consists of premises and a conclusion. An argument is valid if the conclusion follows necessarily from the premises.
Proof Techniques in Logic (Direct, Indirect, Contradiction)
- Direct Proof: Proving a statement by a straightforward chain of reasoning.
- Indirect Proof: Assuming the negation of a statement and deriving a contradiction.
- Proof by Contradiction: Proving a statement by showing that assuming its negation leads to a contradiction.
Normal Forms (Conjunctive and Disjunctive)
- Conjunctive Normal Form (CNF): A proposition is in CNF if it is a conjunction of disjunctions.
- Disjunctive Normal Form (DNF): A proposition is in DNF if it is a disjunction of conjunctions.
Resolution in Propositional and Predicate Logic
Resolution is a proof technique used in both propositional and predicate logic, particularly in automated theorem proving.
Applications of Logic in Computer Science
- Programming: Logic is fundamental in programming languages, especially in conditions and loops.
- Databases: Logical queries and conditions are used in databases.
- Artificial Intelligence: Logic is crucial for reasoning, decision-making, and problem-solving.
Applications of Logic in Mathematics
Logic forms the basis of proofs in mathematics. Mathematical theorems and definitions are written in logical form, and the truth or falsehood of statements is determined using logical deduction.
Differences Between Propositional and Predicate Logic
- Propositional logic deals with simple, declarative propositions and logical connectives.
- Predicate logic allows for the analysis of more complex statements involving variables and quantifiers.
Challenges in Logic
Understanding and mastering logical arguments and proofs can be challenging, especially as complexity increases with the introduction of predicate logic, quantifiers, and functions.
Example: Truth Table for $p \rightarrow q$
Problem: Construct a truth table for $p \rightarrow q$.
Answer:
Step 1: Given Data:
Propositions $p$ and $q$.
Step 2: Solution:
| $p$ | $q$ | $p \rightarrow q$ |
|---|---|---|
| $T$ | $T$ | $T$ |
| $T$ | $F$ | $F$ |
| $F$ | $T$ | $T$ |
| $F$ | $F$ | $T$ |
Step 3: Final Answer:
The truth table shows the possible truth values of $p \rightarrow q$.
Question And Answer Library
Example 1: Propositional Logic – Truth Table
Problem:
Construct a truth table for the proposition $ p \land q $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
- If $ p = T $ and $ q = T $, then $ p \land q = T $.
- If $ p = T $ and $ q = F $, then $ p \land q = F $.
- If $ p = F $ and $ q = T $, then $ p \land q = F $.
- If $ p = F $ and $ q = F $, then $ p \land q = F $.
Step 3: Final Answer:
The truth values for $ p \land q $ are as follows:
- $ p = T, q = T \Rightarrow p \land q = T $
- $ p = T, q = F \Rightarrow p \land q = F $
- $ p = F, q = T \Rightarrow p \land q = F $
- $ p = F, q = F \Rightarrow p \land q = F $
Example 2: Logical Equivalence
Problem:
Prove that $ p \lor \neg p $ is a tautology.
Answer:
Step 1: Given Data:
Proposition $ p $.
Step 2: Solution:
- When $ p = T $, then $ \neg p = F $ and $ p \lor \neg p = T \lor F = T $.
- When $ p = F $, then $ \neg p = T $ and $ p \lor \neg p = F \lor T = T $.
Step 3: Final Answer:
Since $ p \lor \neg p $ is true for all truth values of $ p $, it is a tautology.
Example 3: Modus Ponens
Problem:
Using Modus Ponens, prove that if $ p \rightarrow q $ and $ p $ are true, then $ q $ must also be true.
Answer:
Step 1: Given Data:
- Premise 1: $ p \rightarrow q $
- Premise 2: $ p $
Step 2: Solution:
From Premise 1, we know if $ p $ is true, then $ q $ must also be true.
Since Premise 2 states that $ p $ is true, we can conclude that $ q $ is true.
Step 3: Final Answer:
Therefore, $ q $ is true based on Modus Ponens.
Example 4: De Morgan’s Laws
Problem:
Demonstrate De Morgan’s Law: $ \neg (p \land q) \equiv \neg p \lor \neg q $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
- Consider the case when $ p = T $ and $ q = T $:
$ \neg (T \land T) = \neg T = F $
$ \neg T \lor \neg T = F \lor F = F $ - Consider the case when $ p = T $ and $ q = F $:
$ \neg (T \land F) = \neg F = T $
$ \neg T \lor \neg F = F \lor T = T $ - Consider the case when $ p = F $ and $ q = T $:
$ \neg (F \land T) = \neg F = T $
$ \neg F \lor \neg T = T \lor F = T $ - Consider the case when $ p = F $ and $ q = F $:
$ \neg (F \land F) = \neg F = T $
$ \neg F \lor \neg F = T \lor T = T $
Step 3: Final Answer:
In all cases, $ \neg (p \land q) $ is equivalent to $ \neg p \lor \neg q $.
Example 5: Predicate Logic
Problem:
Translate the statement “All humans are mortal” into predicate logic.
Answer:
Step 1: Given Data:
Let $ H(x) $ represent “x is a human” and $ M(x) $ represent “x is mortal”.
Step 2: Solution:
Using the universal quantifier, we can express this as:
$ \forall x (H(x) \rightarrow M(x)) $
Step 3: Final Answer:
The statement “All humans are mortal” is expressed in predicate logic as $ \forall x (H(x) \rightarrow M(x)) $.
Example 6: Existential Quantifier
Problem:
Translate the statement “There exists a cat that is black” into predicate logic.
Answer:
Step 1: Given Data:
Let $ C(x) $ represent “x is a cat” and $ B(x) $ represent “x is black”.
Step 2: Solution:
Using the existential quantifier, we can express this as:
$ \exists x (C(x) \land B(x)) $
Step 3: Final Answer:
The statement “There exists a cat that is black” is expressed in predicate logic as $ \exists x (C(x) \land B(x)) $.
Example 7: Proof by Contradiction
Problem:
Prove that if $ p $ is true, then $ p \lor q $ is also true using proof by contradiction.
Answer:
Step 1: Given Data:
Assume $ p $ is true.
Step 2: Solution:
Assume for contradiction that $ p \lor q $ is false.
This implies both $ p $ and $ q $ must be false.
However, this contradicts our assumption that $ p $ is true.
Step 3: Final Answer:
Thus, $ p \lor q $ must be true if $ p $ is true.
Example 8: Validity of an Argument
Problem:
Determine if the following argument is valid:
- If it rains, then the ground is wet.
- It is raining.
- Therefore, the ground is wet.
Answer:
Step 1: Given Data:
Premise 1: $ p \rightarrow q $ (If it rains, then the ground is wet)
Premise 2: $ p $ (It is raining)
Step 2: Solution:
Using Modus Ponens, since $ p $ is true, we conclude that $ q $ must also be true.
Step 3: Final Answer:
The argument is valid, and the conclusion is that the ground is wet.
Example 9: Proof Using Modus Tollens
Problem:
Prove the following: If $ p \rightarrow q $ is true and $ \neg q $ is true, then $ \neg p $ must be true.
Answer:
Step 1: Given Data:
Premise 1: $ p \rightarrow q $
Premise 2: $ \neg q $
Step 2: Solution:
From Premise 1, if $ p $ is true, then $ q $ must be true.
Since $ \neg q $ is true, $ p $ cannot be true, leading to $ \neg p $.
Step 3: Final Answer:
Thus, we conclude that $ \neg p $ is true.
Example 10: Logical Equivalence
Problem:
Prove that $ p \lor (q \land r) \equiv (p \lor q) \land (p \lor r) $.
Answer:
Step 1: Given Data:
Propositions $ p $, $ q $, and $ r $.
Step 2: Solution:
- Consider all possible truth values for $ p $, $ q $, and $ r $.
- Check if both expressions yield the same truth values under all combinations.
Step 3: Final Answer:
Since both expressions yield the same truth values, they are logically equivalent.
Example 11: Proving a Predicate Logic Statement
Problem:
Show that $ \forall x (P(x) \lor Q(x)) \rightarrow \exists x P(x) $ is not necessarily true.
Answer:
Step 1: Given Data:
Let $ P(x) $ represent “x is even” and $ Q(x) $ represent “x is odd”.
Step 2: Solution:
If $ \forall x (P(x) \lor Q(x)) $ is true (since every number is either even or odd),
it does not guarantee that $ \exists x P(x) $ is true (e.g., if all x are odd).
Step 3: Final Answer:
The statement is not necessarily true as there can be cases where it fails.
Example 12: Proving a Biconditional
Problem:
Prove that $ p \leftrightarrow q $ is equivalent to $ (p \rightarrow q) \land (q \rightarrow p) $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
- To show biconditional equivalence, prove both directions.
- Assume $ p \leftrightarrow q $ holds true, hence $ p \rightarrow q $ and $ q \rightarrow p $ must also be true.
Step 3: Final Answer:
Thus, $ p \leftrightarrow q $ is equivalent to $ (p \rightarrow q) \land (q \rightarrow p) $.
Example 13: Evaluating Quantifiers
Problem:
Evaluate the truth of the statement: $ \exists x (P(x) \land Q(x)) $ given that $ P(1) $ is true, $ Q(1) $ is false.
Answer:
Step 1: Given Data:
$ P(1) $ is true and $ Q(1) $ is false.
Step 2: Solution:
Since $ Q(1) $ is false, $ P(1) \land Q(1) $ is false.
Step 3: Final Answer:
Thus, $ \exists x (P(x) \land Q(x)) $ is false.
Example 14: Using Universal Quantifiers
Problem:
Show that the statement $ \forall x (P(x) \rightarrow Q(x)) $ holds if $ P(a) $ is false for some specific element $ a $.
Answer:
Step 1: Given Data:
Assume $ P(a) $ is false.
Step 2: Solution:
Since the implication $ P(x) \rightarrow Q(x) $ is true when $ P(x) $ is false,
the statement holds for the specific element $ a $.
Step 3: Final Answer:
Thus, $ \forall x (P(x) \rightarrow Q(x)) $ holds.
Example 15: Implication in Logic
Problem:
Show that the statement $ p \rightarrow (q \rightarrow p) $ is a tautology.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
- Evaluate $ p = T, q = T \Rightarrow T \rightarrow (T \rightarrow T) = T $.
- Evaluate $ p = T, q = F \Rightarrow T \rightarrow (F \rightarrow T) = T $.
- Evaluate $ p = F, q = T \Rightarrow F \rightarrow (T \rightarrow F) = T $.
- Evaluate $ p = F, q = F \Rightarrow F \rightarrow (F \rightarrow F) = T $.
Step 3: Final Answer:
Therefore, $ p \rightarrow (q \rightarrow p) $ is a tautology.
Example 16: Direct Proof
Problem:
Prove that if $ p $ is true, then $ p \lor q $ is true.
Answer:
Step 1: Given Data:
Assume $ p $ is true.
Step 2: Solution:
Since $ p $ is true, regardless of the truth value of $ q $,
it follows that $ p \lor q $ is true.
Step 3: Final Answer:
Thus, $ p \lor q $ is true if $ p $ is true.
Example 17: Proof by Contradiction
Problem:
Prove that $ \sqrt{2} $ is irrational using proof by contradiction.
Answer:
Step 1: Given Data:
Assume, for contradiction, that $ \sqrt{2} $ is rational.
Step 2: Solution:
If $ \sqrt{2} = \frac{a}{b} $ for integers $ a $ and $ b $, then $ 2 = \frac{a^2}{b^2} $ implies $ a^2 = 2b^2 $.
This means $ a^2 $ is even, hence $ a $ must be even. Let $ a = 2k $ for some integer $ k $.
Then $ 4k^2 = 2b^2 $ implies $ b^2 = 2k^2 $, so $ b^2 $ is also even. Thus, both $ a $ and $ b $ are even, contradicting the assumption that they are coprime.
Step 3: Final Answer:
Therefore, $ \sqrt{2} $ is irrational.
Example 18: Using Predicate Logic
Problem:
Translate “For every integer $ n $, $ n^2 \geq 0 $” into predicate logic.
Answer:
Step 1: Given Data:
Let $ P(n) $ represent “$ n^2 \geq 0 $”.
Step 2: Solution:
The statement can be expressed as:
$ \forall n \in \mathbb{Z} (P(n)) $
Step 3: Final Answer:
The predicate logic representation is $ \forall n \in \mathbb{Z} (n^2 \geq 0) $.
Example 19: Negating a Statement
Problem:
Negate the statement “All dogs are friendly.”
Answer:
Step 1: Given Data:
The statement “All dogs are friendly”.
Step 2: Solution:
The negation of the statement is:
“Not all dogs are friendly”, or “There exists at least one dog that is not friendly.”
Step 3: Final Answer:
The negated statement is “There exists at least one dog that is not friendly”.
Example 20: Logical Argument
Problem:
Evaluate the validity of the argument:
- If it is snowing, then it is cold.
- It is not cold.
- Therefore, it is not snowing.
Answer:
Step 1: Given Data:
Premise 1: $ p \rightarrow q $ (If it is snowing, then it is cold)
Premise 2: $ \neg q $ (It is not cold)
Step 2: Solution:
Using Modus Tollens, from $ p \rightarrow q $ and $ \neg q $, we can conclude $ \neg p $ (it is not snowing).
Step 3: Final Answer:
The argument is valid, and the conclusion is that it is not snowing.
Example 21: Evaluating a Predicate
Problem:
Evaluate the truth of the predicate $ P(x): x^2 > 4 $ for $ x = 3 $.
Answer:
Step 1: Given Data:
Let $ x = 3 $.
Step 2: Solution:
Evaluate $ P(3): 3^2 > 4 \Rightarrow 9 > 4 $.
Step 3: Final Answer:
The predicate $ P(3) $ is true.
Example 22: Demonstrating Universal Quantification
Problem:
Show that $ \forall x (x + 2 > x) $ is true for all integers.
Answer:
Step 1: Given Data:
Let $ x $ be any integer.
Step 2: Solution:
- For any integer $ x $, $ x + 2 > x $ holds true.
- The additional 2 ensures that the left side is always greater than the right side.
Step 3: Final Answer:
Therefore, $ \forall x (x + 2 > x) $ is true for all integers.
Example 23: Existential Quantification
Problem:
Show that $ \exists x (x^2 = 4) $ is true.
Answer:
Step 1: Given Data:
We need to find an $ x $ such that $ x^2 = 4 $.
Step 2: Solution:
- For $ x = 2 $, $ 2^2 = 4 $ is true.
- For $ x = -2 $, $ (-2)^2 = 4 $ is also true.
Step 3: Final Answer:
Thus, $ \exists x (x^2 = 4) $ is true.
Example 24: Proof of a Biconditional Statement
Problem:
Prove that $ p \leftrightarrow q $ is equivalent to $ (p \rightarrow q) \land (q \rightarrow p) $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
To prove equivalence, we need to show that both sides yield the same truth values.
Assume $ p \leftrightarrow q $ is true, then both $ p \rightarrow q $ and $ q \rightarrow p $ must be true.
Step 3: Final Answer:
Thus, $ p \leftrightarrow q $ is equivalent to $ (p \rightarrow q) \land (q \rightarrow p) $.
Example 25: Proving a Logical Statement
Problem:
Prove that $ p \lor (p \land q) \equiv p $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
Using the definition of logical OR:
- If $ p $ is true, then $ p \lor (p \land q) = T$ regardless of $ q $.
- If $ p $ is false, then $ p \land q $ is false, hence $ p \lor (p \land q) = F$.
Step 3: Final Answer:
Therefore, $ p \lor (p \land q) \equiv p $ is proven.
Example 26: Proving an Argument
Problem:
Evaluate the argument:
- If the light is on, then the switch is up.
- The light is not on.
- Therefore, the switch is down.
Answer:
Step 1: Given Data:
Premise 1: $ p \rightarrow q $ (If the light is on, then the switch is up)
Premise 2: $ \neg p $ (The light is not on)
Step 2: Solution:
From Premise 1 and Premise 2, we cannot conclude anything about the state of the switch. The argument is invalid.
Step 3: Final Answer:
Thus, the conclusion does not necessarily follow from the premises.
Example 27: Proving an Existential Statement
Problem:
Show that $ \exists x (P(x)) $ is true if $ P(1) $ is true.
Answer:
Step 1: Given Data:
Assume $ P(1) $ is true.
Step 2: Solution:
Since there exists at least one $ x $ (specifically $ x = 1 $) such that $ P(x) $ is true.
Step 3: Final Answer:
Therefore, $ \exists x (P(x)) $ is true.
Example 28: Understanding Logical Connectives
Problem:
Explain the logical connective $ p \land \neg q $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
- The expression $ p \land \neg q $ means that $ p $ must be true and $ q $ must be false.
- This connective is true only when $ p $ is true and $ q $ is false.
Step 3: Final Answer:
Thus, $ p \land \neg q $ is true if and only if $ p $ is true and $ q $ is false.
Example 29: Validating a Quantified Statement
Problem:
Evaluate the statement $ \forall x (x + 1 > x) $.
Answer:
Step 1: Given Data:
Let $ x $ be any real number.
Step 2: Solution:
For any real number $ x $, the statement $ x + 1 > x $ holds true because adding 1 increases the value.
Step 3: Final Answer:
Thus, $ \forall x (x + 1 > x) $ is true.
Example 30: Evaluating a Predicate Logic Statement
Problem:
Translate the statement “There exists a number that is even” into predicate logic.
Answer:
Step 1: Given Data:
Let $ E(x) $ represent “$ x $ is even”.
Step 2: Solution:
The statement can be expressed as:
$ \exists x E(x) $
Step 3: Final Answer:
The predicate logic representation is $ \exists x (E(x)) $.
Example 31: Proof by Contradiction
Problem:
Prove that if $ p \land q $ is true, then both $ p $ and $ q $ are true.
Answer:
Step 1: Given Data:
Assume $ p \land q $ is true.
Step 2: Solution:
By the definition of conjunction, for $ p \land q $ to be true, both $ p $ and $ q $ must be true.
Step 3: Final Answer:
Therefore, $ p $ is true and $ q $ is true.
Example 32: Validity of a Quantified Argument
Problem:
Evaluate the argument:
- For all $ x $, if $ P(x) $ then $ Q(x) $ is true.
- $ P(a) $ is true.
- Therefore, $ Q(a) $ is true.
Answer:
Step 1: Given Data:
Premise 1: $ \forall x (P(x) \rightarrow Q(x)) $
Premise 2: $ P(a) $
Step 2: Solution:
From the premises, since $ P(a) $ is true and $ \forall x (P(x) \rightarrow Q(x)) $ holds, it follows that $ Q(a) $ must also be true.
Step 3: Final Answer:
Thus, the argument is valid, and the conclusion is that $ Q(a) $ is true.
Example 33: Understanding a Logical Implication
Problem:
Explain the meaning of the implication $ p \rightarrow q $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
The implication $ p \rightarrow q $ states that if $ p $ is true, then $ q $ must also be true.
It is only false when $ p $ is true and $ q $ is false.
Step 3: Final Answer:
Thus, $ p \rightarrow q $ expresses a conditional relationship between $ p $ and $ q $.
Example 34: Logical Equivalence and Truth
Problem:
Prove that $ \neg (p \land q) \equiv \neg p \lor \neg q $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
Using the truth table approach:
- For all combinations of truth values of $ p $ and $ q $, both expressions yield the same truth value.
Step 3: Final Answer:
Thus, $ \neg (p \land q) \equiv \neg p \lor \neg q $ is proven.
Example 35: Proof Using Universal Instantiation
Problem:
Prove that if $ \forall x P(x) $, then $ P(2) $ is true.
Answer:
Step 1: Given Data:
Assume $ \forall x P(x) $ is true.
Step 2: Solution:
By universal instantiation, we can conclude that $ P(2) $ must also be true.
Step 3: Final Answer:
Thus, $ P(2) $ is true.
Example 36: Evaluating an Existential Statement
Problem:
Show that $ \exists x (x^2 = 9) $ is true.
Answer:
Step 1: Given Data:
We need to find an $ x $ such that $ x^2 = 9 $.
Step 2: Solution:
- For $ x = 3 $, $ 3^2 = 9 $ is true.
- For $ x = -3 $, $ (-3)^2 = 9 $ is also true.
Step 3: Final Answer:
Thus, $ \exists x (x^2 = 9) $ is true.
Example 37: Showing an Argument’s Validity
Problem:
Evaluate the argument:
- If it is sunny, then we will go to the park.
- It is sunny.
- Therefore, we will go to the park.
Answer:
Step 1: Given Data:
Premise 1: $ p \rightarrow q $ (If it is sunny, then we will go to the park)
Premise 2: $ p $ (It is sunny)
Step 2: Solution:
Using Modus Ponens, since $ p $ is true, we conclude that $ q $ must also be true.
Step 3: Final Answer:
Thus, we will go to the park.
Example 38: Proving a Conditional Statement
Problem:
Prove that $ p \rightarrow (q \lor r) $ is valid when $ p $ is true.
Answer:
Step 1: Given Data:
Assume $ p $ is true.
Step 2: Solution:
If $ p $ is true, $ p \rightarrow (q \lor r) $ is true regardless of the truth values of $ q $ and $ r $.
Step 3: Final Answer:
Thus, the statement $ p \rightarrow (q \lor r) $ is valid.
Example 39: Understanding Biconditional Statements
Problem:
Explain the biconditional statement $ p \leftrightarrow q $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
The statement $ p \leftrightarrow q $ means that both $ p $ and $ q $ must have the same truth value.
It is true if both are true or both are false.
Step 3: Final Answer:
Therefore, $ p \leftrightarrow q $ establishes an equivalence between the truth values of $ p $ and $ q $.
Example 40: Using Logical Connectives
Problem:
Evaluate the statement $ p \lor (q \land r) $ when $ p = F $, $ q = T $, and $ r = F $.
Answer:
Step 1: Given Data:
Let $ p = F $, $ q = T $, and $ r = F $.
Step 2: Solution:
Evaluate:
$ q \land r = T \land F = F $
Then, $ p \lor (q \land r) = F \lor F = F $.
Step 3: Final Answer:
Thus, the statement evaluates to $ F $.
Example 41: Understanding Truth Values
Problem:
Determine the truth value of the statement $ (p \lor q) \land (p \rightarrow r) $ given $ p = T $, $ q = F $, and $ r = F $.
Answer:
Step 1: Given Data:
Let $ p = T $, $ q = F $, and $ r = F $.
Step 2: Solution:
Evaluate:
$ p \lor q = T \lor F = T $
$ p \rightarrow r = T \rightarrow F = F $
Then, $ (p \lor q) \land (p \rightarrow r) = T \land F = F $.
Step 3: Final Answer:
Thus, the truth value of the statement is $ F $.
Example 42: Negating a Quantified Statement
Problem:
Negate the statement “For all integers $ n $, $ n^2 \geq 0 $” in predicate logic.
Answer:
Step 1: Given Data:
The statement is $ \forall n (n^2 \geq 0) $.
Step 2: Solution:
The negation is:
$ \neg \forall n (n^2 \geq 0) \equiv \exists n (n^2 < 0) $.
Step 3: Final Answer:
The negated statement is $ \exists n (n^2 < 0) $.
Example 43: Validating Logical Statements
Problem:
Show that $ p \rightarrow (q \land r) $ is equivalent to $ (p \rightarrow q) \land (p \rightarrow r) $.
Answer:
Step 1: Given Data:
Propositions $ p $, $ q $, and $ r $.
Step 2: Solution:
- Use a truth table to evaluate both expressions.
- Check if both yield the same truth values for all combinations of $ p $, $ q $, and $ r $.
Step 3: Final Answer:
Since both expressions yield the same truth values, they are equivalent.
Example 44: Evaluating Quantifiers in a Statement
Problem:
Show that $ \forall x (P(x) \lor Q(x)) \rightarrow \exists x P(x) $ is not valid.
Answer:
Step 1: Given Data:
Let $ P(x) $ be “x is even” and $ Q(x) $ be “x is odd”.
Step 2: Solution:
The premise states that for every integer $ x $, either $ x $ is even or odd.
However, it does not guarantee that there exists an integer $ x $ such that $ P(x) $ is true.
Step 3: Final Answer:
Thus, the statement is not valid.
Example 45: Establishing Logical Equivalence
Problem:
Prove that $ (p \lor q) \land r \equiv (p \land r) \lor (q \land r) $.
Answer:
Step 1: Given Data:
Propositions $ p $, $ q $, and $ r $.
Step 2: Solution:
Using the distributive property of logic, we can rewrite:
$ (p \lor q) \land r = (p \land r) \lor (q \land r) $.
Step 3: Final Answer:
Thus, the logical equivalence is proven.
Example 46: Validating a Predicate Logic Statement
Problem:
Translate “Every student has a laptop” into predicate logic.
Answer:
Step 1: Given Data:
Let $ S(x) $ represent “$ x $ is a student” and $ L(x) $ represent “$ x $ has a laptop”.
Step 2: Solution:
The statement can be expressed as:
$ \forall x (S(x) \rightarrow L(x)) $
Step 3: Final Answer:
The predicate logic representation is $ \forall x (S(x) \rightarrow L(x)) $.
Example 47: Proving a Statement with Universal Quantifiers
Problem:
Show that if $ \forall x P(x) $ is true, then $ P(5) $ is true.
Answer:
Step 1: Given Data:
Assume $ \forall x P(x) $ is true.
Step 2: Solution:
By universal instantiation, it follows that $ P(5) $ must also be true.
Step 3: Final Answer:
Thus, $ P(5) $ is true.
Example 48: Proving a Biconditional
Problem:
Prove that $ p \leftrightarrow q $ is equivalent to $ (p \rightarrow q) \land (q \rightarrow p) $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
To show equivalence, both sides must yield the same truth values.
Assume $ p \leftrightarrow q $ is true; therefore, both implications must also be true.
Step 3: Final Answer:
Thus, $ p \leftrightarrow q $ is equivalent to $ (p \rightarrow q) \land (q \rightarrow p) $.
Example 49: Proof Using Modus Tollens
Problem:
Prove that if $ p \rightarrow q $ is true and $ \neg q $ is true, then $ \neg p $ must also be true.
Answer:
Step 1: Given Data:
Premise 1: $ p \rightarrow q $
Premise 2: $ \neg q $
Step 2: Solution:
From Premise 1, if $ p $ is true, then $ q $ must also be true.
Since $ \neg q $ is true, it follows that $ p $ cannot be true, leading to $ \neg p $.
Step 3: Final Answer:
Therefore, $ \neg p $ is true.
Example 50: Evaluating an Argument
Problem:
Evaluate the argument:
- If it rains, then the ground is wet.
- The ground is not wet.
- Therefore, it is not raining.
Answer:
Step 1: Given Data:
Premise 1: $ p \rightarrow q $ (If it rains, then the ground is wet)
Premise 2: $ \neg q $ (The ground is not wet)
Step 2: Solution:
Using Modus Tollens, since $ \neg q $ is true, we conclude $ \neg p $ (it is not raining).
Step 3: Final Answer:
Thus, the argument is valid, and the conclusion is that it is not raining.
Example 51: Truth Value of a Conditional Statement
Problem:
Determine the truth value of the statement $ p \rightarrow q $ given $ p = F $ and $ q = T $.
Answer:
Step 1: Given Data:
Let $ p = F $ and $ q = T $.
Step 2: Solution:
Evaluate $ p \rightarrow q$:
If $ p $ is false, then $ p \rightarrow q $ is true regardless of the truth value of $ q $.
Step 3: Final Answer:
Thus, the truth value of the statement $ p \rightarrow q $ is $ T $.
Example 52: Satisfiability of a Predicate
Problem:
Is the predicate $ P(x): x^2 < 0 $ satisfiable?
Answer:
Step 1: Given Data:
Consider the predicate $ P(x) $ for real numbers.
Step 2: Solution:
Since $ x^2 $ is always non-negative for real numbers, there are no values of $ x $ such that $ x^2 < 0 $.
Step 3: Final Answer:
Thus, the predicate $ P(x) $ is unsatisfiable.
Example 53: Evaluating an Existential Statement
Problem:
Show that $ \exists x (x^2 = -1) $ is not true for real numbers.
Answer:
Step 1: Given Data:
We need to find an $ x $ such that $ x^2 = -1 $.
Step 2: Solution:
No real number squared results in a negative number, thus no $ x $ satisfies this equation.
Step 3: Final Answer:
Therefore, $ \exists x (x^2 = -1) $ is false for real numbers.
Example 54: Validating Logical Connectives
Problem:
Evaluate the logical connective $ \neg (p \lor q) $.
Answer:
Step 1: Given Data:
Assume $ p $ and $ q $ can have any truth values.
Step 2: Solution:
Using De Morgan’s laws:
$ \neg (p \lor q) \equiv \neg p \land \neg q $.
Step 3: Final Answer:
Thus, $ \neg (p \lor q) $ is equivalent to $ \neg p \land \neg q $.
Example 55: Understanding the Biconditional
Problem:
Explain the meaning of the statement $ p \leftrightarrow q $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
The biconditional $ p \leftrightarrow q $ means that both propositions have the same truth value.
It is true if both $ p $ and $ q $ are true or both are false.
Step 3: Final Answer:
Therefore, $ p \leftrightarrow q $ establishes equivalence between the truth values of $ p $ and $ q $.
Example 56: Demonstrating Universal Quantification
Problem:
Show that the statement $ \forall x (x + 1 > x) $ is true for all integers.
Answer:
Step 1: Given Data:
Let $ x $ be any integer.
Step 2: Solution:
For any integer $ x $, the statement $ x + 1 > x $ holds true because adding 1 increases the value.
Step 3: Final Answer:
Thus, $ \forall x (x + 1 > x) $ is true for all integers.
Example 57: Evaluating an Argument
Problem:
Evaluate the argument:
- If it is sunny, then we will go to the beach.
- It is not sunny.
- Therefore, we will not go to the beach.
Answer:
Step 1: Given Data:
Premise 1: $ p \rightarrow q $ (If it is sunny, then we will go to the beach)
Premise 2: $ \neg p $ (It is not sunny)
Step 2: Solution:
From the premises, we cannot conclude anything about going to the beach; the argument is invalid.
Step 3: Final Answer:
Thus, the conclusion does not necessarily follow from the premises.
Example 58: Using Predicate Logic
Problem:
Translate “There exists a cat that is black” into predicate logic.
Answer:
Step 1: Given Data:
Let $ C(x) $ represent “x is a cat” and $ B(x) $ represent “x is black”.
Step 2: Solution:
The statement can be expressed as:
$ \exists x (C(x) \land B(x)) $
Step 3: Final Answer:
The predicate logic representation is $ \exists x (C(x) \land B(x)) $.
Example 59: Validating a Logical Statement
Problem:
Show that $ p \rightarrow (q \lor r) $ is true when $ p $ is true.
Answer:
Step 1: Given Data:
Assume $ p $ is true.
Step 2: Solution:
If $ p $ is true, $ p \rightarrow (q \lor r) $ is true regardless of the truth values of $ q $ and $ r $.
Step 3: Final Answer:
Thus, the statement is true.
Example 60: Understanding Implication
Problem:
Explain the logical implication $ p \rightarrow q $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
The implication $ p \rightarrow q $ states that if $ p $ is true, then $ q $ must also be true.
It is only false when $ p $ is true and $ q $ is false.
Step 3: Final Answer:
Thus, $ p \rightarrow q $ expresses a conditional relationship between $ p $ and $ q $.
Example 61: Evaluating a Predicate Logic Statement
Problem:
Show that $ \exists x (P(x) \lor Q(x)) $ is true if either $ P(a) $ or $ Q(a) $ is true.
Answer:
Step 1: Given Data:
Assume $ P(a) $ is true or $ Q(a) $ is true.
Step 2: Solution:
If either condition holds, then $ \exists x (P(x) \lor Q(x)) $ must be satisfied for at least one $ x $.
Step 3: Final Answer:
Therefore, $ \exists x (P(x) \lor Q(x)) $ is true.
Example 62: Evaluating an Argument’s Validity
Problem:
Evaluate the argument:
- If it is raining, then the ground is wet.
- The ground is wet.
- Therefore, it is raining.
Answer:
Step 1: Given Data:
Premise 1: $ p \rightarrow q $ (If it is raining, then the ground is wet)
Premise 2: $ q $ (The ground is wet)
Step 2: Solution:
From the premises, we cannot conclude that $ p $ is true; this argument is invalid.
Step 3: Final Answer:
Thus, the conclusion does not necessarily follow from the premises.
Example 63: Validating a Quantified Statement
Problem:
Evaluate the statement $ \forall x (P(x) \rightarrow Q(x)) $ given $ P(2) $ is true.
Answer:
Step 1: Given Data:
Assume $ P(2) $ is true.
Step 2: Solution:
This implies that $ Q(2) $ must also be true under the universal quantification assumption.
Step 3: Final Answer:
Thus, the statement $ \forall x (P(x) \rightarrow Q(x)) $ holds.
Example 64: Logical Equivalence
Problem:
Prove that $ p \rightarrow (q \land r) \equiv (p \rightarrow q) \land (p \rightarrow r) $.
Answer:
Step 1: Given Data:
Propositions $ p $, $ q $, and $ r $.
Step 2: Solution:
Using truth values, check both expressions.
If both yield the same values for all cases, they are equivalent.
Step 3: Final Answer:
Thus, the equivalence is proven.
Example 65: Evaluating Logical Connectives
Problem:
Evaluate the expression $ \neg (p \land q) $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
Using De Morgan’s law, we can rewrite:
$ \neg (p \land q) \equiv \neg p \lor \neg q $.
Step 3: Final Answer:
Thus, the expression simplifies to $ \neg p \lor \neg q $.
Example 66: Truth Table for a Compound Statement
Problem:
Construct a truth table for $ (p \land q) \lor r $.
Answer:
Step 1: Given Data:
Propositions $ p $, $ q $, and $ r $.
Step 2: Solution:
- Evaluate for all combinations of $ p $, $ q $, and $ r $:
- If $ p = T, q = T, r = T$, then $ (T \land T) \lor T = T $.
- If $ p = T, q = T, r = F$, then $ (T \land T) \lor F = T $.
- If $ p = T, q = F, r = T$, then $ (T \land F) \lor T = T $.
- If $ p = T, q = F, r = F$, then $ (T \land F) \lor F = F $.
- Continue evaluating for $ F $ values of $ p $ and $ q $.
Step 3: Final Answer:
The truth table will display the results for all combinations of truth values.
Example 67: Evaluating Quantified Statements
Problem:
Show that $ \exists x (P(x) \land Q(x)) $ is false if both $ P(a) $ and $ Q(a) $ are false.
Answer:
Step 1: Given Data:
Assume $ P(a) $ is false and $ Q(a) $ is false.
Step 2: Solution:
Since both predicates are false, $ P(a) \land Q(a) $ is also false, leading to the conclusion that there is no $ x $ satisfying the statement.
Step 3: Final Answer:
Thus, $ \exists x (P(x) \land Q(x)) $ is false.
Example 68: Logical Arguments and Conclusions
Problem:
Evaluate the argument:
- If $ p $, then $ q $.
- If $ q $, then $ r $.
- Therefore, if $ p $, then $ r $.
Answer:
Step 1: Given Data:
Premise 1: $ p \rightarrow q $
Premise 2: $ q \rightarrow r $
Step 2: Solution:
Using hypothetical syllogism, we conclude $ p \rightarrow r $ holds.
Step 3: Final Answer:
Thus, the argument is valid.
Example 69: Logical Connective Evaluation
Problem:
Evaluate the statement $ \neg (p \lor q) $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
Using De Morgan’s laws:
$ \neg (p \lor q) \equiv \neg p \land \neg q $.
Step 3: Final Answer:
Thus, the expression simplifies to $ \neg p \land \neg q $.
Example 70: Logical Equivalence Proof
Problem:
Prove that $ p \land (q \lor r) \equiv (p \land q) \lor (p \land r) $.
Answer:
Step 1: Given Data:
Propositions $ p $, $ q $, and $ r $.
Step 2: Solution:
Using the distributive property, we can rewrite:
$ p \land (q \lor r) = (p \land q) \lor (p \land r) $.
Step 3: Final Answer:
Thus, the logical equivalence is proven.
Example 71: Predicate Logic Example
Problem:
Translate the statement “Some cats are black” into predicate logic.
Answer:
Step 1: Given Data:
Let $ C(x) $ represent “$ x $ is a cat” and $ B(x) $ represent “$ x $ is black”.
Step 2: Solution:
The statement can be expressed as:
$ \exists x (C(x) \land B(x)) $
Step 3: Final Answer:
The predicate logic representation is $ \exists x (C(x) \land B(x)) $.
Example 72: Proof of Validity
Problem:
Prove that the statement $ p \rightarrow (q \rightarrow p) $ is a tautology.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
- When $ p = T $, $ p \rightarrow (q \rightarrow p) $ is true regardless of $ q $.
- When $ p = F $, $ p \rightarrow (q \rightarrow p) $ is true as well.
Step 3: Final Answer:
Therefore, $ p \rightarrow (q \rightarrow p) $ is a tautology.
Example 73: Understanding Negation
Problem:
Negate the statement “All dogs are friendly.”
Answer:
Step 1: Given Data:
The statement is $ \forall x (D(x) \rightarrow F(x)) $, where $ D(x) $ means “x is a dog” and $ F(x) $ means “x is friendly”.
Step 2: Solution:
The negation is:
$ \neg \forall x (D(x) \rightarrow F(x)) \equiv \exists x (D(x) \land \neg F(x)) $.
Step 3: Final Answer:
The negated statement is “There exists a dog that is not friendly”.
Example 74: Evaluating Logical Statements
Problem:
Evaluate the logical statement $ p \land (q \lor r) $ when $ p = T $, $ q = F $, and $ r = T $.
Answer:
Step 1: Given Data:
Let $ p = T $, $ q = F $, and $ r = T $.
Step 2: Solution:
Evaluate:
$ q \lor r = F \lor T = T $
Then, $ p \land (q \lor r) = T \land T = T $.
Step 3: Final Answer:
Thus, the statement evaluates to $ T $.
Example 75: Proving a Predicate Logic Statement
Problem:
Translate the statement “For every integer $ n $, $ n^2 \geq 0 $” into predicate logic.
Answer:
Step 1: Given Data:
Let $ P(n) $ represent “$ n^2 \geq 0 $”.
Step 2: Solution:
The statement can be expressed as:
$ \forall n (P(n)) $
Step 3: Final Answer:
The predicate logic representation is $ \forall n (n^2 \geq 0) $.
Example 76: Validating a Quantified Statement
Problem:
Show that if $ P(a) $ is true, then $ \exists x P(x) $ is true.
Answer:
Step 1: Given Data:
Assume $ P(a) $ is true for some specific $ a $.
Step 2: Solution:
Since $ P(a) $ holds for that particular $ a $, it follows that there exists an $ x $ such that $ P(x) $ is true.
Step 3: Final Answer:
Therefore, $ \exists x P(x) $ is true.
Example 77: Understanding Logical Implication
Problem:
Explain the implication $ p \rightarrow q $ in words.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
The implication $ p \rightarrow q $ can be read as “if $ p $ is true, then $ q $ is also true.”
It is false only when $ p $ is true and $ q $ is false.
Step 3: Final Answer:
Thus, the implication states a conditional relationship between $ p $ and $ q $.
Example 78: Proof Using Predicate Logic
Problem:
Translate “There exists a student who passed the exam” into predicate logic.
Answer:
Step 1: Given Data:
Let $ S(x) $ represent “$ x $ is a student” and $ P(x) $ represent “$ x $ passed the exam”.
Step 2: Solution:
The statement can be expressed as:
$ \exists x (S(x) \land P(x)) $
Step 3: Final Answer:
The predicate logic representation is $ \exists x (S(x) \land P(x)) $.
Example 79: Evaluating an Argument
Problem:
Evaluate the argument:
- If the door is locked, then we cannot enter.
- The door is not locked.
- Therefore, we can enter.
Answer:
Step 1: Given Data:
Premise 1: $ p \rightarrow q $ (If the door is locked, then we cannot enter)
Premise 2: $ \neg p $ (The door is not locked)
Step 2: Solution:
From the premises, we cannot conclude anything about entering; the argument is invalid.
Step 3: Final Answer:
Thus, the conclusion does not necessarily follow from the premises.
Example 80: Evaluating a Predicate Statement
Problem:
Show that $ \forall x (x^2 \geq 0) $ is true for all integers.
Answer:
Step 1: Given Data:
Let $ x $ be any integer.
Step 2: Solution:
For any integer $ x $, the statement $ x^2 \geq 0 $ holds true because squaring any integer results in a non-negative value.
Step 3: Final Answer:
Thus, $ \forall x (x^2 \geq 0) $ is true for all integers.
Example 81: Logical Connective Evaluation
Problem:
Evaluate the statement $ p \land (q \lor r) $ when $ p = F $, $ q = T $, and $ r = T $.
Answer:
Step 1: Given Data:
Let $ p = F $, $ q = T $, and $ r = T $.
Step 2: Solution:
Evaluate:
$ q \lor r = T \lor T = T $
Then, $ p \land (q \lor r) = F \land T = F $.
Step 3: Final Answer:
Thus, the statement evaluates to $ F $.
Example 82: Understanding Tautologies
Problem:
Identify if $ p \lor \neg p $ is a tautology.
Answer:
Step 1: Given Data:
Proposition $ p $.
Step 2: Solution:
For any truth value of $ p $:
- If $ p = T $, then $ p \lor \neg p = T \lor F = T $.
- If $ p = F $, then $ p \lor \neg p = F \lor T = T $.
Step 3: Final Answer:
Since $ p \lor \neg p $ is true for all values, it is a tautology.
Example 83: Demonstrating Quantified Statements
Problem:
Prove the statement $ \forall x (x + 0 = x) $ for all integers.
Answer:
Step 1: Given Data:
Let $ x $ be any integer.
Step 2: Solution:
For any integer $ x $, adding 0 to it does not change its value: $ x + 0 = x $.
Step 3: Final Answer:
Thus, the statement $ \forall x (x + 0 = x) $ is true.
Example 84: Using Modus Tollens
Problem:
Use Modus Tollens to prove that if $ p \rightarrow q $ is true and $ \neg q $ is true, then $ \neg p $ must be true.
Answer:
Step 1: Given Data:
Premise 1: $ p \rightarrow q $
Premise 2: $ \neg q $
Step 2: Solution:
From Premise 1, if $ p $ is true, then $ q $ must be true.
Since $ \neg q $ is true, $ p $ cannot be true, leading to $ \neg p $.
Step 3: Final Answer:
Therefore, $ \neg p $ is true.
Example 85: Evaluating Quantifiers
Problem:
Evaluate the truth of the statement $ \exists x (x^2 < 0) $.
Answer:
Step 1: Given Data:
Consider the predicate $ P(x) $ where $ P(x) $ is “$ x^2 < 0 $”.
Step 2: Solution:
Since no real number squared is negative, $ P(x) $ cannot be satisfied for any $ x $.
Step 3: Final Answer:
Thus, the statement $ \exists x (x^2 < 0) $ is false.
Example 86: Validating Arguments
Problem:
Evaluate the argument:
- If the alarm is on, then the door is locked.
- The door is not locked.
- Therefore, the alarm is off.
Answer:
Step 1: Given Data:
Premise 1: $ p \rightarrow q $ (If the alarm is on, then the door is locked)
Premise 2: $ \neg q $ (The door is not locked)
Step 2: Solution:
From Modus Tollens, since $ \neg q $ is true, we conclude $ \neg p $ (the alarm is off).
Step 3: Final Answer:
Thus, the argument is valid, and the conclusion is that the alarm is off.
Example 87: Understanding Implication
Problem:
Explain the meaning of the implication $ q \rightarrow p $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
The implication $ q \rightarrow p $ states that if $ q $ is true, then $ p $ must also be true.
It is only false when $ q $ is true and $ p $ is false.
Step 3: Final Answer:
Thus, the implication establishes a conditional relationship between $ q $ and $ p $.
Example 88: Truth Table for $ p \land q $
Problem:
Construct a truth table for the proposition $ p \land q $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
Evaluate the truth values:
- If $ p = T $ and $ q = T $, then $ p \land q = T $.
- If $ p = T $ and $ q = F $, then $ p \land q = F $.
- If $ p = F $ and $ q = T $, then $ p \land q = F $.
- If $ p = F $ and $ q = F $, then $ p \land q = F $.
Step 3: Final Answer:
Thus, the truth values for $ p \land q $ are established based on the combinations.
Example 89: Proving a Statement
Problem:
Prove that $ p \lor (q \land r) \equiv (p \lor q) \land (p \lor r) $.
Answer:
Step 1: Given Data:
Propositions $ p $, $ q $, and $ r $.
Step 2: Solution:
Using distribution, we can rewrite:
$ p \lor (q \land r) = (p \lor q) \land (p \lor r) $.
Step 3: Final Answer:
Thus, the logical equivalence is proven.
Example 90: Understanding Quantifiers
Problem:
Evaluate the truth of the statement $ \forall x (x + 1 \geq x) $.
Answer:
Step 1: Given Data:
Let $ x $ be any real number.
Step 2: Solution:
For any real number $ x $, $ x + 1 \geq x $ holds true since adding 1 increases the value.
Step 3: Final Answer:
Thus, $ \forall x (x + 1 \geq x) $ is true.
Example 91: Proving a Conditional Statement
Problem:
Show that if $ p $ is true, then $ p \rightarrow (q \lor r) $ is also true.
Answer:
Step 1: Given Data:
Assume $ p $ is true.
Step 2: Solution:
Since $ p $ is true, $ p \rightarrow (q \lor r) $ is true regardless of the truth values of $ q $ and $ r $.
Step 3: Final Answer:
Thus, the statement is true.
Example 92: Understanding a Logical Argument
Problem:
Evaluate the argument:
- If the sky is clear, then we will see the stars.
- The sky is clear.
- Therefore, we will see the stars.
Answer:
Step 1: Given Data:
Premise 1: $ p \rightarrow q $ (If the sky is clear, then we will see the stars)
Premise 2: $ p $ (The sky is clear)
Step 2: Solution:
Using Modus Ponens, since $ p $ is true, we conclude that $ q $ must also be true.
Step 3: Final Answer:
Thus, we will see the stars.
Example 93: Logical Equivalence
Problem:
Prove that $ \neg (p \lor q) \equiv \neg p \land \neg q $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
Using De Morgan’s laws:
$ \neg (p \lor q) \equiv \neg p \land \neg q $.
Step 3: Final Answer:
Thus, the logical equivalence is proven.
Example 94: Proving an Argument’s Validity
Problem:
Evaluate the argument:
- If it snows, then the roads are slippery.
- The roads are not slippery.
- Therefore, it does not snow.
Answer:
Step 1: Given Data:
Premise 1: $ p \rightarrow q $ (If it snows, then the roads are slippery)
Premise 2: $ \neg q $ (The roads are not slippery)
Step 2: Solution:
Using Modus Tollens, since $ \neg q $ is true, we conclude $ \neg p $ (it does not snow).
Step 3: Final Answer:
Thus, the argument is valid, and the conclusion is that it does not snow.
Example 95: Evaluating a Logical Connective
Problem:
Evaluate the statement $ p \rightarrow (q \land r) $ when $ p = F $, $ q = T $, and $ r = F $.
Answer:
Step 1: Given Data:
Let $ p = F $, $ q = T $, and $ r = F $.
Step 2: Solution:
Evaluate $ q \land r = T \land F = F $
Then, $ p \rightarrow (q \land r) = F \rightarrow F = T $.
Step 3: Final Answer:
Thus, the statement evaluates to $ T $.
Example 96: Understanding Quantifiers
Problem:
Evaluate the statement $ \exists x (x + 1 = 0) $ for integers.
Answer:
Step 1: Given Data:
Consider the predicate $ P(x) $ where $ P(x) $ means “$ x + 1 = 0 $”.
Step 2: Solution:
The only integer that satisfies this is $ x = -1 $.
Step 3: Final Answer:
Thus, the statement $ \exists x (x + 1 = 0) $ is true.
Example 97: Proof of Validity Using Implication
Problem:
Prove that $ p \rightarrow (q \rightarrow p) $ is a tautology.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
- When $ p = T $, $ p \rightarrow (q \rightarrow p) = T \rightarrow (q \rightarrow T) = T $.
- When $ p = F $, the statement $ p \rightarrow (q \rightarrow p) $ becomes $ F \rightarrow (q \rightarrow F) = T $.
Step 3: Final Answer:
Therefore, $ p \rightarrow (q \rightarrow p) $ is a tautology.
Example 98: Validating Quantified Statements
Problem:
Evaluate the statement $ \forall x (x^2 \geq 0) $ for all real numbers.
Answer:
Step 1: Given Data:
Let $ x $ be any real number.
Step 2: Solution:
For any real number $ x $, $ x^2 \geq 0 $ holds true because squaring a number results in a non-negative value.
Step 3: Final Answer:
Thus, $ \forall x (x^2 \geq 0) $ is true for all real numbers.
Example 99: Understanding Logical Connectives
Problem:
Evaluate the logical statement $ \neg (p \land q) $.
Answer:
Step 1: Given Data:
Propositions $ p $ and $ q $.
Step 2: Solution:
Using De Morgan’s laws:
$ \neg (p \land q) \equiv \neg p \lor \neg q $.
Step 3: Final Answer:
Thus, the expression simplifies to $ \neg p \lor \neg q $.
Example 100: Using Logical Implication
Problem:
Explain the implications of the statement $ p \lor q \rightarrow r $.
Answer:
Step 1: Given Data:
Propositions $ p $, $ q $, and $ r $.
Step 2: Solution:
The statement $ p \lor q \rightarrow r $ implies that if either $ p $ or $ q $ is true, then $ r $ must also be true.
It is false only when $ p \lor q $ is true and $ r $ is false.
Step 3: Final Answer:
Thus, the implication establishes a conditional relationship between the truth values of $ p \lor q $ and $ r $.
4o mini
