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Chapter 1: Problem 2

Show that the following simple, but very useful relations, which are widely used in mathematical reasoning, are true: a) \(\neg(A \wedge B) \Leftrightarrow \neg A \vee \neg B\); b) \(\neg(A \vee B) \Leftrightarrow \neg A \wedge \neg B\); c) \((A \Rightarrow B) \Leftrightarrow(\neg B \Rightarrow \neg A)\); d) \((A \Rightarrow B) \Leftrightarrow(\neg A \vee B)\) e) \(\neg(A \Rightarrow B) \Leftrightarrow A \wedge \neg B\).

Short Answer

Expert verified
Question: Prove the validity of the following logical equivalences using truth tables: a) A 鈭� (A 鈭� B) 鈮� A b) A 鈭� (A 鈭� B) 鈮� A c) A 鈭� (卢A 鈭� B) 鈮� A 鈭� B d) 卢(A 鈭� B) 鈮� 卢A 鈭� 卢B e) 卢(A 鈭� B) 鈮� 卢A 鈭� 卢B Answer: We can show the validity of these logical equivalences using truth tables, which allow us to consider all possible combinations of truth values for the propositions A and B. If both sides of the equivalence have the same truth values for all combinations, the equivalence is valid. We start by creating a truth table for each part (a) to (e), evaluating both sides of the equivalence for each combination of truth values for A and B. After evaluating all parts with truth tables, we can conclude that each equivalence is valid because both sides of each equivalence have the same truth values for all possible combinations of truth values for A and B.

Step by step solution

01

Truth Table for (a)

Begin by listing all possible combinations of truth values for propositions A and B, which are four: (T, T), (T, F), (F, T), and (F, F). Then, evaluate both sides of the equivalence for each combination. If both sides have the same truth values for all combinations, the equivalence will be proven.
02

Truth Table for (b)

Similarly, create a truth table for part b, evaluating both sides of the equivalence for each combination of truth values for A and B.
03

Truth Table for (c)

For part c, create a truth table evaluating both sides of the equivalence for each combination of truth values for A and B.
04

Truth Table for (d)

Create a truth table for part d, evaluating both sides of the equivalence for each combination of truth values for A and B.
05

Truth Table for (e)

Lastly, create a truth table for part e, evaluating both sides of the equivalence for each combination of truth values for A and B. After evaluating all five parts with truth tables, we can conclude that each equivalence is true because both sides of each equivalence have the same truth values for all possible combinations of truth values for A and B.

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Key Concepts

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

Truth Table
Truth tables are invaluable tools in understanding and proving logical equivalences. They are grids that list all possible truth values of the involved propositions and the resulting truth values of the logical expressions.
For our exercise, we have two propositions, A and B, leading to four possible combinations of truth values:
  • (True, True)
  • (True, False)
  • (False, True)
  • (False, False)
For each part of the exercise, a truth table allows us to evaluate the logical expressions under these conditions, ensuring that both sides of each equivalence produce the same outcome.
This method makes it straightforward to visually interpret the connection and verify if an equivalence holds. Once you have all combinations lined up, you simply compare the columns of each expression. If both columns match completely, the logical equivalence is true.
Logical Reasoning
Logical reasoning is a critical skill in solving logical equivalences. It involves drawing conclusions based on given propositions and determining the validity and truth of the logical statements.
In our exercise, logical reasoning helps us understand the relationships defined by each equivalence. For instance, to know why \(eg(A \wedge B) \Leftrightarrow eg A \vee eg B\) is true, we utilize reasoning skills.
Here, it becomes clear that if \(A\) and \(B\) are both not true, not both their negations will ensure the equivalence to hold. Thus, logical reasoning guides you in not only constructing truth tables but also in interpreting them.
This ensures you understand why a certain logical equivalence holds regardless of merely seeing a truth table that confirms its validity.
Propositional Logic
Propositional logic, at its core, deals with propositions that can either be true or false. It forms the basis for a majority of logical expressions covered in this exercise. Understanding propositional logic helps in breaking down complex logical statements into simpler components.
Each part of the given exercise is built upon basic propositions \(A\) and \(B\), and the relationships between them are formed using logical operators like conjunction (\(\wedge\)), disjunction (\(\vee\)), and implication (\(\Rightarrow\)).
By mastering propositional logic, you can tackle more intricate logical expressions and build solid arguments. For example, from knowing that \(eg A \wedge eg B\) implies neither \(A\) nor \(B\) is true, you develop a robust ability to formalize and analyze arguments in all areas of logic and mathematics.
Thus, having a firm grasp of propositional logic not only aids in this exercise but enhances logical thinking overall.

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Most popular questions from this chapter

In Exercises 1,2 , and 3 the letters \(A, B\), and \(C\) denote subsets of a set \(M\). \- Verify the following relations. a) \((A \subset C) \wedge(B \subset C) \Leftrightarrow((A \cup B) \subset C)\); b) \((C \subset A) \wedge(C \subset B) \Leftrightarrow(C \subset(A \cap B))\); c) \(C_{M}\left(C_{M} A\right)=A\); d) \(\left(A \subset C_{M} B\right) \Leftrightarrow\left(B \subset C_{M} A\right)\) e) \((A \subset B) \Leftrightarrow\left(C_{M} A \supset C_{M} B\right)\).

We shall deal only with sets. Since a set consisting of different elements may itself be an element of another set, logicians usually denote all sets by uppercase letters. In the present exercise, it is very convenient to do so. a) Verify that the statement $$ \forall x \exists y \forall z(z \in y \Leftrightarrow \exists w(z \in w \wedge w \in x)) $$ expresses the axiom of union, according to which \(y\) is the union of the sets belonging to \(x\). b) State which axioms of set theory are represented by the following statements: $$ \begin{aligned} &\forall x \forall y \forall z((z \in x \Leftrightarrow z \in y) \Leftrightarrow x=y) \\ &\forall x \forall y \exists z \forall v(v \in z \Leftrightarrow(v=x \vee v=y)) \\ &\forall x \exists y \forall z(z \in y \Leftrightarrow \forall u(u \in z \Rightarrow u \in x)) \\ &\exists x(\forall y(\neg \exists z(z \in y) \Rightarrow y \in x) \wedge \\ &\quad \wedge \forall w(w \in x \Rightarrow \forall u(\forall v(v \in u \Leftrightarrow(v=w \vee v \in w)) \Rightarrow u \in x))) \end{aligned} $$ c) Verify that the formula $$ \begin{aligned} \forall z &\left(z \in f \Rightarrow\left(\exists x_{1} \exists y_{1}\left(x_{1} \in x \wedge y_{1} \in y \wedge z=\left(x_{1}, y_{1}\right)\right)\right)\right) \wedge \\ & \wedge \forall x_{1}\left(x_{1} \in x \Rightarrow \exists y_{1} \exists z\left(y_{1} \in y \wedge z=\left(x_{1}, y_{1}\right) \wedge z \in f\right)\right) \wedge \\ & \wedge \forall x_{1} \forall y_{1} \forall y_{2}\left(\exists z_{1} \exists z_{2}\left(z_{1} \in f \wedge z_{2} \in f \wedge z_{1}=\left(x_{1}, y_{1}\right) \wedge z_{2}=\left(x_{1}, y_{2}\right)\right) \Rightarrow\right.\\\ &\left.\Rightarrow y_{1}=y_{2}\right) \end{aligned} $$ imposes three successive restrictions on the set \(f: f\) is a subset of \(x \times y ;\) the projection of \(f\) on \(x\) is equal to \(x\); to each element \(x_{1}\) of \(x\) there corresponds exactly one \(y_{1}\) in \(y\) such that \(\left(x_{1}, y_{1}\right) \in f\). Thus what we have here is a definition of a mapping \(f: x \rightarrow y\). This example shows yet again that the formal expression of a statement is by no means always the shortest and most transparent in comparison with its expression in ordinary language. Taking this circumstance into account, we shall henceforth use logical symbolism only to the extent that it seems useful to us to achieve greater compactness or clarity of exposition.

Prove the following statements. a) \(A \cup(B \cup C)=(A \cup B) \cup C=: A \cup B \cup C\); b) \(A \cap(B \cap C)=(A \cap B) \cap C=: A \cap B \cap C\); c) \(A \cap(B \cup C)=(A \cap B) \cup(A \cap C)\); d) \(A \cup(B \cap C)=(A \cup B) \cap(A \cup C)\).

The composition \(\mathcal{R}_{2} \circ \mathcal{R}_{1}\) of the relations \(\mathcal{R}_{1}\) and \(\mathcal{R}_{2}\) is defined as follows: $$ \mathcal{R}_{2} \circ \mathcal{R}_{1}:=\left\\{(x, z) \mid \exists y\left(x \mathcal{R}_{1} y \wedge y \mathcal{R}_{2} z\right)\right\\} $$ In particular, if \(\mathcal{R}_{1} \subset X \times Y\) and \(\mathcal{R}_{2} \subset Y \times Z\), then \(\mathcal{R}=\mathcal{R}_{2} \circ \mathcal{R}_{1} \subset X \times Z\), and $$ x \mathcal{R} z:=\exists y\left((y \in Y) \wedge\left(x \mathcal{R}_{1} y\right) \wedge\left(y \mathcal{R}_{2} z\right)\right) $$ a) Let \(\Delta_{X}\) be the diagonal of \(X^{2}\) and \(\Delta_{Y}\) the diagonal of \(Y^{2}\). Show that if the relations \(\mathcal{R}_{1} \subset X \times Y\) and \(\mathcal{R}_{2} \subset Y \times X\) are such that \(\left(\mathcal{R}_{2} \circ \mathcal{R}_{1}=\Delta_{X}\right) \wedge\left(\mathcal{R}_{1} \circ \mathcal{R}_{2}=\right.\) \(\left.\Delta_{Y}\right)\), then both relations are functional and define mutually inverse mappings of \(X\) and \(Y\). b) Let \(\mathcal{R} \subset X^{2}\). Show that the condition of transitivity of the relation \(\mathcal{R}\) is equivalent to the condition \(\mathcal{R} \circ \mathcal{R} \subset \mathcal{R}\). c) The relation \(\mathcal{R}^{\prime} \subset Y \times X\) is called the transpose of the relation \(\mathcal{R} \subset X \times Y\) if \(\left(y \mathcal{R}^{\prime} x\right) \Leftrightarrow(x \mathcal{R} y)\) Show that a relation \(\mathcal{R} \subset X^{2}\) is antisymmetric if and only if \(\mathcal{R} \cap \mathcal{R}^{\prime} \subset \Delta_{X}\) d) Verify that any two elements of \(X\) are connected (in some order) by the relation \(\mathcal{R} \subset X^{2}\) if and only if \(\mathcal{R} \cup \mathcal{R}^{\prime}=X^{2}\)

Verify that the following statements about a mapping \(f: X \rightarrow Y\) are equivalent: a) \(f\) is injective; b) \(f^{-1}(f(A))=A\) for every \(A \subset X\); c) \(f(A \cap B)=f(A) \cap f(B)\) for any two subsets \(A\) and \(B\) of \(X\); d) \(f(A) \cap f(B)=\varnothing \Leftrightarrow A \cap B=\varnothing\); e) \(f(A \backslash B)=f(A) \backslash f(B)\) whenever \(X \supset A \supset B\).
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