Step 1: Formal Definition of Subset and Set Difference

To prove that $C - B \subseteq C - A$ given the condition $A \subseteq B$, we must rely on the fundamental definitions of set theory:

• Subset Definition: $A \subseteq B$ if and only if $x \in A \implies x \in B$ [Per the definition of subset inclusion].
• Set Difference Definition: $C - S = \{x : x \in C \text{ and } x \notin S\}$ [Per the definition of relative complement].

Step 2: Logical Derivation of the Inclusion

To prove the inclusion $C - B \subseteq C - A$, we must demonstrate that any arbitrary element $x$ belonging to $C - B$ also necessarily belongs to $C - A$.

Let $x$ be an arbitrary element such that:

$x \in (C - B)$

By the definition of set difference, this implies:

$x \in C$ and $x \notin B$ [Condition 1]

Step 3: Applying the Contrapositive of the Given Condition

We are given $A \subseteq B$. Applying the logical law of contraposition:

$A \subseteq B \iff (x \in A \implies x \in B)$

The contrapositive is:

$(x \notin B \implies x \notin A)$ [By the Rule of Contrapositive in propositional logic]

Step 4: Synthesizing the Argument

Since we established from Condition 1 that $x \notin B$, we can now invoke the result from Step 3:

1. We know $x \in C$ [From Step 2].
2. We know $x \notin B$, which implies $x \notin A$ [From Step 3].

Combining these, we have $x \in C$ and $x \notin A$.

By the definition of set difference, this exactly satisfies the condition for membership in $C - A$:

$x \in (C - A)$

Step 5: Visual Representation

Final Conclusion

Since every element $x \in (C - B)$ has been shown to be an element of $(C - A)$, by the definition of a subset, we conclude:

Final Solution: $C - B \subseteq C - A$

Step 1: Analyzing the Logical Structure

The statement is: If $A \subset B$ and $B \in C$, then $A \in C$.

To evaluate this, we define the relations as follows:

• $A \subset B$: Every element $x$ such that $x \in A$ implies $x \in B$.
• $B \in C$: The set $B$ is an element of the set $C$.
• Conclusion: $A \in C$: The set $A$ must be an element of the set $C$.

[Per the definitions of Subset and Elementhood in Set Theory, the relation of "being an element of" is not necessarily transitive across the relation of "being a subset of".]

Step 2: Determining Truth Value

The statement is False. Inclusion ($A \subset B$) defines a relationship between sets, while elementhood ($B \in C$) defines membership of a set within a collection. Having $B$ as a member of $C$ does not imply that every subset of $B$ is also a member of $C$.

Step 3: Providing a Counter-Example

To disprove the universal quantification of this statement, we need only one counter-example where the premises are true but the conclusion is false.

Verification of Premises:

• $A \subset B$: Since $1 \in \{1\}$ and $1 \in \{1, 2\}$, the inclusion holds. [By Subset Definition].
• $B \in C$: The set $\{1, 2\}$ is indeed the unique element of set $C$. The premise holds.

Verification of Conclusion:

• $A \in C$: Is $\{1\} \in \{\{1, 2\}\}$? No. The only element in $C$ is $\{1, 2\}$. Therefore, $A \notin C$.

Step 4: Visualizing the Membership Hierarchy

Final Solution: The statement is False. As shown by the counter-example where $A=\{1\}$, $B=\{1, 2\}$, and $C=\{\{1, 2\}\}$, it is observed that while $A \subset B$ and $B \in C$, it does not follow that $A \in C$.

Step 1: Logical Framework of Equivalence

To prove that the condition $A - B = \phi$ is equivalent to the fundamental subset relation $A \subseteq B$, we must demonstrate that $A - B = \phi \implies A \subseteq B$ and $A \subseteq B \implies A - B = \phi$.

Step 2: Proof of $A - B = \phi \implies A \subseteq B$

Assume $A - B = \phi$. We utilize the definition of set difference: $A - B = \{x : x \in A \text{ and } x \notin B\}$.

• If $A - B = \phi$, then there exists no element $x$ such that $x \in A$ and $x \notin B$ [Definition of the Empty Set].
• Suppose, for the sake of contradiction, that $A \not\subseteq B$. This implies there exists at least one element $y \in A$ such that $y \notin B$.
• By the definition of set difference, $y \in A - B$.
• However, since $A - B = \phi$, this implies $y \in \phi$, which contradicts the definition of the empty set [Axiom of Emptiness].
• Therefore, our assumption that $A \not\subseteq B$ must be false. Thus, $A \subseteq B$.

Step 3: Proof of $A \subseteq B \implies A - B = \phi$

Assume $A \subseteq B$. We wish to prove $A - B = \phi$.

• Let $x \in A - B$. By definition, $x \in A$ and $x \notin B$ [Set Difference Property].
• Given $A \subseteq B$, if $x \in A$, then it necessarily follows that $x \in B$ [Definition of Subset].
• We have reached a logical contradiction: $x \in B$ and $x \notin B$.
• Since the existence of an element $x \in A - B$ leads to a contradiction, the set $A - B$ must contain no elements.
• Therefore, $A - B = \phi$.

Visual Representation of $A \subseteq B$

Analytical Summary Table

Final Solution: The condition $A - B = \phi$ is logically equivalent to $A \subseteq B$, as both directions of implication hold under the axioms of set theory.

Logical Analysis & Theoretical Foundation

The statement posits that the subset relation is transitive. Given three sets $A$, $B$, and $C$, if $A$ is a subset of $B$ ($A \subseteq B$) and $B$ is a subset of $C$ ($B \subseteq C$), then it must hold that $A$ is a subset of $C$ ($A \subseteq C$).

Step 1: Formal Definition of Subset Relation

• $A \subseteq B$ implies that for all elements $x$, if $x \in A$, then $x \in B$. [Definition of Subset]
• $B \subseteq C$ implies that for all elements $x$, if $x \in B$, then $x \in C$. [Definition of Subset]

Step 2: Logical Derivation via Transitive Property

To prove $A \subseteq C$, we must show that for any arbitrary element $x \in A$, it necessarily follows that $x \in C$.

1. Let $x$ be an arbitrary element such that $x \in A$.
2. Since it is given that $A \subseteq B$, and $x \in A$, it follows by the definition of a subset that $x \in B$.
3. Since it is given that $B \subseteq C$, and we have established $x \in B$, it follows by the definition of a subset that $x \in C$.
4. By the Law of Syllogism (transitivity of implication), if $x \in A \implies x \in B$ and $x \in B \implies x \in C$, then $x \in A \implies x \in C$.

Step 3: Visual Representation

Step 4: Comparison Table of Subset Logic

Final Solution: The statement is True. By the definition of the subset relation, the property is transitive: for any sets $A, B,$ and $C$, if $A \subseteq B$ and $B \subseteq C$, then $A \subseteq C$.

Statement Evaluation

The statement is False.

Theoretical Justification

To disprove the universal quantification of the statement "If $x \in A$ and $A \in B$, then $x \in B$", we must provide a counter-example where the antecedent ($x \in A \land A \in B$) is true, but the consequent ($x \in B$) is false. This demonstrates that the inclusion relationship between elements and sets does not possess the property of transitivity in the manner suggested by the statement [Per the definition of Set Membership $\in$].

Counter-Example Construction

Let us define sets $A$ and $B$ such that $x$ is an element of $A$, and $A$ itself is an element of $B$.

• Let $x = 1$.
• Let $A = \{1\}$. Note that $1 \in A$ is true.
• Let $B = \{ \{1\}, 2 \}$. Here, $A \in B$ is true because the set $\{1\}$ is an element of $B$.

Now, we evaluate the consequent $x \in B$:

The elements of $B$ are $\{1\}$ and $2$. The element $x$ (which is $1$) is not contained within $B$ as a direct member. Although the set containing $1$ is in $B$, the element $1$ is not. Therefore, $1 \notin B$.

Visual Representation

Logical Analysis

The failure of this statement arises because set membership ($\in$) is not transitive. If $x \in A$ and $A \in B$, $x$ is an element of an element of $B$, which does not imply $x$ is an element of $B$ itself. The correct transitive property would involve the subset relation ($\subseteq$): If $x \in A$ and $A \subseteq B$, then $x \in B$.

Final Solution: The statement is False. Example: If $A = \{1\}$ and $B = \{ \{1\}, 2 \}$, then $1 \in A$ and $A \in B$, but $1 \notin B$.

Theoretical Foundation & Logical Equivalence

To demonstrate that a condition $A \cup B = B$ is equivalent to $A \subseteq B$, we must prove the bidirectional implication (biconditional statement): $A \subseteq B \iff A \cup B = B$. This relies on the Axioms of Set Theory, specifically the definitions of Union and Subset relations.

Step 1: Forward Implication ($\implies$) - Proving $A \subseteq B \implies A \cup B = B$

Assume $A \subseteq B$. We aim to show that $A \cup B = B$ by demonstrating mutual inclusion ($A \cup B \subseteq B$ and $B \subseteq A \cup B$).

• Part A: Show $A \cup B \subseteq B$
  Let $x \in A \cup B$. By the definition of union, $x \in A$ or $x \in B$ [Definition of Set Union].
  Case 1: If $x \in A$, then since $A \subseteq B$, it follows that $x \in B$ [Definition of Subset].
  Case 2: If $x \in B$, the condition is trivially satisfied.
  Thus, $x \in B$ in all cases, confirming $A \cup B \subseteq B$.
• Part B: Show $B \subseteq A \cup B$
  For any set $B$, it is a fundamental property that any element belonging to $B$ also belongs to the union of $B$ with any set $A$ [Identity/Monotonicity of Union].
  Formally, if $y \in B$, then $y \in B$ or $y \in A$, which implies $y \in A \cup B$.
  Thus, $B \subseteq A \cup B$.

By the Axiom of Extensionality, since $A \cup B \subseteq B$ and $B \subseteq A \cup B$, we conclude $A \cup B = B$.

Step 2: Reverse Implication ($\impliedby$) - Proving $A \cup B = B \implies A \subseteq B$

Assume $A \cup B = B$. We aim to show $A \subseteq B$.

• Let $x$ be an arbitrary element such that $x \in A$.
• By the property of union, $x \in A \cup B$ [Definition of Union].
• Since we are given that $A \cup B = B$, it follows by substitution that $x \in B$.
• Since $x \in A$ implies $x \in B$, it follows by definition that $A \subseteq B$ [Definition of Subset].

Step 3: Visual Representation (Venn Diagram)

Summary of Logic

Final Solution: Since the forward and reverse implications hold true, it is rigorously established that $A \cup B = B$ is logically equivalent to $A \subseteq B$.

Step 1: Theoretical Foundation of Set Equivalence

To demonstrate that the condition $A \cap B = A$ is equivalent to the fundamental inclusion relation of subsets, we must analyze the logical implications of the intersection operation under the axioms of Set Theory [Per Zermelo-Fraenkel Set Theory axioms].

Step 2: Logical Derivation of the Condition

We analyze the statement $A \cap B = A$ using the definition of set intersection: $A \cap B = \{x : x \in A \text{ and } x \in B\}$.

• Forward Implication ($\implies$): Assume $A \cap B = A$. By the definition of intersection, for any element $x \in (A \cap B)$, it must be that $x \in A$ and $x \in B$. Since our assumption states $A \cap B = A$, any element $x \in A$ must also be an element of $A \cap B$. Therefore, if $x \in A$, then $x \in (A \cap B)$, which implies $x \in A$ and $x \in B$. Consequently, $x \in B$. Thus, $A \subseteq B$.
• Backward Implication ($\impliedby$): Assume $A \subseteq B$. To show $A \cap B = A$, we must demonstrate both $A \cap B \subseteq A$ and $A \subseteq A \cap B$.
  • By the definition of intersection, $A \cap B \subseteq A$ is always true [By the properties of set operations].
  • If $A \subseteq B$, then for any element $x \in A$, it follows that $x \in B$. Since $x \in A$ and $x \in B$, it follows that $x \in (A \cap B)$. Thus, $A \subseteq A \cap B$.
  • Since $A \cap B \subseteq A$ and $A \subseteq A \cap B$, by the principle of double inclusion, $A \cap B = A$.

Step 3: Visual Representation of the Equivalence

The following SVG illustrates the scenario where $A \subseteq B$, resulting in $A \cap B = A$.

Step 4: Formal Summary of Equivalent Conditions

The condition $A \cap B = A$ holds if and only if $A$ is a subset of $B$. This relationship is fundamentally defined by the behavior of elements residing in the intersection of two sets where the constraints of the subset relation are satisfied.

Final Solution: The condition $A \cap B = A$ is logically equivalent to $A \subseteq B$.

Step 1: Determining the Elements of Set $A$

To identify the elements of set $A$, we solve the quadratic equation $x^2 - 8x + 12 = 0$. [By the Quadratic Formula or Factorization Method].

Factoring the quadratic expression:

$x^2 - 6x - 2x + 12 = 0$

$x(x - 6) - 2(x - 6) = 0$

$(x - 2)(x - 6) = 0$

Setting each factor to zero, we obtain the roots $x = 2$ and $x = 6$.

Thus, $A = \{2, 6\}$.

Step 2: Listing the Sets for Comparison

Based on our calculation and the provided problem statement:

• $A = \{2, 6\}$
• $B = \{2, 4, 6\}$
• $C = \{2, 4, 6, 8, \dots \}$ (The set of all even natural numbers)
• $D = \{6\}$

Step 3: Evaluating Subset Relationships

A set $X$ is a subset of $Y$ ($X \subseteq Y$) if and only if every element of $X$ is also an element of $Y$. [Per the Definition of Subset].

Step 4: Geometric Representation of Set Inclusion

Final Solution: The subset relationships are as follows: $D \subset A \subset B \subset C$. Additionally, $D \subset B$, $D \subset C$, and $A \subset C$.

Step 1: Establishing the Logical Framework

To prove $B = C$, we must demonstrate that $B \subseteq C$ and $C \subseteq B$ [Per the Axiom of Extensionality]. We utilize the distributive laws of set theory to manipulate the given equations:

Given (i): $A \cup B = A \cup C$

Given (ii): $A \cap B = A \cap C$

Step 2: Proving $B \subseteq C$

Let $x$ be an arbitrary element such that $x \in B$.

• Since $x \in B$, it follows that $x \in A \cup B$ [By the definition of Union].
• From Given (i), $A \cup B = A \cup C$, therefore $x \in A \cup C$.
• This implies $x \in A$ or $x \in C$ [By the definition of Union].

We analyze these two cases:

1. Case 1: $x \in A$. Since we assumed $x \in B$ and now $x \in A$, it follows that $x \in A \cap B$ [By the definition of Intersection]. Using Given (ii), $A \cap B = A \cap C$, therefore $x \in A \cap C$. This implies $x \in C$.
2. Case 2: $x \in C$. The condition is satisfied directly.

In both cases, $x \in C$. Thus, $B \subseteq C$.

Step 3: Proving $C \subseteq B$

Let $y$ be an arbitrary element such that $y \in C$.

• Since $y \in C$, it follows that $y \in A \cup C$.
• From Given (i), $A \cup C = A \cup B$, therefore $y \in A \cup B$.
• This implies $y \in A$ or $y \in B$.

We analyze these two cases:

1. Case 1: $y \in A$. Since we assumed $y \in C$ and now $y \in A$, it follows that $y \in A \cap C$. Using Given (ii), $A \cap C = A \cap B$, therefore $y \in A \cap B$. This implies $y \in B$.
2. Case 2: $y \in B$. The condition is satisfied directly.

In both cases, $y \in B$. Thus, $C \subseteq B$.

Step 4: Logical Synthesis

Since we have established $B \subseteq C$ and $C \subseteq B$, by the definition of set equality, we conclude $B = C$ [By the Principle of Bi-conditional Inclusion].

Final Solution: $B = C$

Logical Analysis of the Proposition

The proposition states: If $A \subset B$ and $x \notin B$, then $x \notin A$.

Step 1: Identifying the Logical Structure

Let $P$ be the condition $A \subset B$.
Let $Q$ be the condition $x \notin B$.
The conclusion to be tested is $R$, which is $x \notin A$.
The logical form is: $(A \subset B \land x \notin B) \implies (x \notin A)$.

Step 2: Theoretical Justification using Set Membership Definitions

By definition, a set $A$ is a subset of $B$ ($A \subset B$) if and only if for every element $y$, $y \in A \implies y \in B$ [Per the Axiom of Extensionality and Definition of Subset].

The contrapositive of the implication $y \in A \implies y \in B$ is $y \notin B \implies y \notin A$ [By the Law of Contraposition in Classical Logic].

Step 3: Verification of the Statement

Since the definition of $A \subset B$ necessitates that every element in $A$ must also be in $B$, it is logically impossible for an element $x$ to be in $A$ while being absent from $B$. If $x$ were in $A$, it would necessarily be in $B$ [By the Definition of Subset].

Therefore, if $x$ is not in $B$ ($x \notin B$), $x$ cannot be in $A$ ($x \notin A$).

Step 4: Formal Proof by Contradiction

1. Assume the premise is true: $A \subset B$ and $x \notin B$.
2. Suppose the conclusion is false: $x \in A$.
3. Since $A \subset B$, if $x \in A$, then $x$ must be in $B$ [By Definition of Subset].
4. This results in $x \in B$, which contradicts our initial premise that $x \notin B$.
5. Because the assumption $x \in A$ leads to a contradiction, the negation $x \notin A$ must be true.

Visual Representation of Logical Inclusion

Final Solution: The statement is True. By the definition of a subset ($A \subset B$), all elements of $A$ are contained within $B$. Thus, if an element $x$ is not in $B$, it cannot be an element of $A$, as that would violate the definition of $A$ being a subset of $B$.

Step 1: Formal Logical Definition

To prove the identity $A \cup (A \cap B) = A$, we employ the Distributive Law of sets, which states that for any sets $A, B,$ and $C$, the union operation distributes over the intersection operation [Per the Distributive Law: $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$].

Step 2: Applying the Distributive Law

Starting with the left-hand side (LHS) of the equation:

$LHS = A \cup (A \cap B)$

Applying the Distributive Law ($A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$) where $C = B$:

$A \cup (A \cap B) = (A \cup A) \cap (A \cup B)$

Step 3: Simplification via Idempotent and Domination Laws

We apply the Idempotent Law, which states that $A \cup A = A$ [Per Idempotent Law]. Substituting this into our expression:

$(A) \cap (A \cup B)$

Step 4: Applying Absorption Property

We analyze the term $(A) \cap (A \cup B)$. Since $A \subseteq (A \cup B)$ is a fundamental property of sets [Per the subset definition where $x \in A \implies x \in (A \cup B)$], any set intersected with its superset results in the set itself.

Specifically: $A \cap (A \cup B) = A$ [Per the Absorption Law].

Step 5: Visual Representation (Venn Diagram)

The shaded region represents the intersection $A \cap B$. Taking the union of set $A$ (the entire left circle) with this shaded region ($A \cap B$) results in the entire circle $A$.

Final Solution: $A \cup (A \cap B) = A$. The identity is proven.

Step 1: Theoretical Framework

To prove the identity $A \cap (A \cup B) = A$, we utilize the fundamental axioms of Set Theory, specifically the Distributive Law and the Identity/Absorption properties. We will provide both a logical derivation using set properties and a rigorous element-wise proof.

Step 2: Algebraic Derivation using Set Laws

We begin with the expression $A \cap (A \cup B)$. We apply the Distributive Law of intersection over union, which states that $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$.

Applying this law where $X=A$, $Y=A$, and $Z=B$:

$A \cap (A \cup B) = (A \cap A) \cup (A \cap B)$ [Distributive Law]

By the Idempotent Law, we know that $A \cap A = A$:

$A \cap (A \cup B) = A \cup (A \cap B)$

Finally, we invoke the Absorption Law, which states that for any sets $A$ and $B$, $A \cup (A \cap B) = A$:

$A \cap (A \cup B) = A$

Step 3: Element-wise Logical Proof

To prove $A \cap (A \cup B) = A$, we must show that $A \cap (A \cup B) \subseteq A$ and $A \subseteq A \cap (A \cup B)$.

• Part I: Show $A \cap (A \cup B) \subseteq A$
  Let $x \in A \cap (A \cup B)$. By definition of intersection, $x \in A$ AND $x \in (A \cup B)$. Since $x \in A$ is explicitly stated, it follows immediately that $A \cap (A \cup B) \subseteq A$.
• Part II: Show $A \subseteq A \cap (A \cup B)$
  Let $x \in A$. By the definition of union, if $x \in A$, then $x \in (A \cup B)$ must also be true. Since $x \in A$ and $x \in (A \cup B)$, by the definition of intersection, $x \in A \cap (A \cup B)$. Thus, $A \subseteq A \cap (A \cup B)$.

Step 4: Visual Representation (Venn Diagram)

Final Solution: Through the application of the Distributive Law and the Absorption Law, it is rigorously established that $A \cap (A \cup B) = A$.

Step 1: Logical Analysis of the Statement

The given statement is: If $x \in A$ and $A \not\subset B$, then $x \in B$.

In symbolic logic, we examine the hypothesis:

• $x \in A$: Element $x$ is contained within set $A$.
• $A \not\subset B$: Set $A$ is not a subset of set $B$, implying there exists at least one element $y$ such that $y \in A$ and $y \notin B$ [Definition of Improper Subset].

Step 2: Constructing a Counterexample

To prove the statement is false, we only need to identify one instance where the hypothesis is true but the conclusion ($x \in B$) is false.

Step 3: Verification of Conditions

Let us verify the conditions against the proposed sets:

• Check $x \in A$: Since $A = \{1, 2\}$, the element $1 \in A$ is true.
• Check $A \not\subset B$: Since $1 \in A$ but $1 \notin B$, the condition $A \not\subset B$ is true [By definition of Subset Exclusion].
• Check Conclusion $x \in B$: Our chosen $x$ is $1$. Since $B = \{2, 3\}$, it is clear that $1 \notin B$.

Step 4: Final Deduction

The statement is False. The inclusion of an element in set $A$ does not guarantee its inclusion in set $B$ simply because $A$ is not a subset of $B$; the definition of "not a subset" merely requires the existence of at least one element in $A$ that is absent from $B$. If $x$ happens to be that specific element, the statement fails.

Final Solution: The statement is False. Counterexample: Let $A = \{1, 2\}$, $B = \{2, 3\}$, and $x = 1$. Here $x \in A$ and $A \not\subset B$, but $x \notin B$.

Theoretical Foundation & Logical Equivalence

To demonstrate that a condition $A \subseteq B$ is equivalent to its subsequent logical implications, we utilize the Principle of Set Inclusion and the definition of the Power Set. Equivalence is established by showing that if one statement is true, all others must necessarily follow.

Step 1: Formal Definition of Subset

By definition, $A \subseteq B$ implies that every element $x$ which is a member of set $A$ is necessarily a member of set $B$. Formally:
$A \subseteq B \iff \forall x, (x \in A \implies x \in B)$ [Definition of Subset].

Step 2: Analytical Derivation of Equivalence

To show that $A \subseteq B$ is equivalent to other standard set-theoretic conditions (such as $A - B = \emptyset$ or $A \cup B = B$), we examine the logical flow:

• Implication $A \subseteq B \implies A - B = \emptyset$: If $A \subseteq B$, then no element exists in $A$ that is not also in $B$. Thus, the set of elements in $A$ but not in $B$ must be the null set $\emptyset$.
• Implication $A - B = \emptyset \implies A \subseteq B$: If $A - B = \emptyset$, then there are no elements $x$ such that $x \in A$ and $x \notin B$. This confirms that for all $x \in A$, $x$ must be in $B$.

Step 3: Geometric Visualization of Subset Inclusion

The following SVG illustrates the inclusion of set $A$ within the universal boundary of set $B$ using Venn notation.

Step 4: Summary of Logical Equivalences

Since each condition can be derived from the definition of a subset and leads back to the existence of $A$ within $B$ [By the Law of Set Equality], the conditions are proven to be equivalent.

Final Solution: The condition $A \subseteq B$ is equivalent to $A \cup B = B$, $A \cap B = A$, and $A - B = \emptyset$, as each statement logically necessitates that all elements of $A$ reside within $B$.

Step 1: Logical Formalization of the Statement

The statement is: If $A \not\subset B$ and $B \not\subset C$, then $A \not\subset C$.

To determine the validity of this implication, we must test whether the conclusion $A \not\subset C$ necessarily follows from the premises $A \not\subset B$ and $B \not\subset C$. If there exists even one counterexample where the premises are true but the conclusion is false, the statement is classified as false.

Step 2: Searching for a Counterexample

We look for sets $A, B,$ and $C$ such that:

• $A \not\subset B$ is true.
• $B \not\subset C$ is true.
• $A \subset C$ is true (which would negate the conclusion $A \not\subset C$).

Let us define the following sets:

• $A = \{1, 2\}$
• $B = \{2, 3\}$
• $C = \{1, 2, 4\}$

Step 3: Verification of Conditions

Step 4: Visual Representation via Venn Diagram

Step 5: Conclusion

Since we have provided a specific case ($A=\{1,2\}, B=\{2,3\}, C=\{1,2,4\}$) where the premises are satisfied but the conclusion $A \not\subset C$ fails (because $A \subset C$), the statement is logically invalid.

Final Solution: The statement is False. Example: $A = \{1, 2\}, B = \{2, 3\}, C = \{1, 2, 4\}$. Here, $A \not\subset B$ and $B \not\subset C$, but $A \subset C$.

Initial Theoretical Setup

To satisfy the conditions $A \cap B \neq \phi$, $B \cap C \neq \phi$, $A \cap C \neq \phi$, and $A \cap B \cap C = \phi$, we must construct sets such that every pair of sets shares at least one common element, but there is no element common to all three sets simultaneously. This configuration represents a Venn diagram where three circles overlap in pairs but do not possess a central common region.

Step 1: Assigning Elements to Pairwise Intersections

Define the sets based on distinct elements to ensure no triple intersection occurs. We assign specific elements to the regions $A \cap B$, $B \cap C$, and $A \cap C$ as follows:

• Let element $x \in (A \cap B)$ such that $x \notin C$.
• Let element $y \in (B \cap C)$ such that $y \notin A$.
• Let element $z \in (A \cap C)$ such that $z \notin B$.

Step 2: Constructing the Sets

By distributing these elements into sets $A$, $B$, and $C$, we formulate the following sets:

• $A = \{x, z\}$
• $B = \{x, y\}$
• $C = \{y, z\}$

Step 3: Verification of Conditions

Geometric Visualization

Final Solution: The sets are $A = \{1, 2\}$, $B = \{1, 3\}$, and $C = \{2, 3\}$, where $A \cap B = \{1\}$, $B \cap C = \{3\}$, $A \cap C = \{2\}$, and $A \cap B \cap C = \phi$.

Step 1: Establishing the Initial Identities

Given the hint and the properties of sets, we utilize the identity relating sets to their unions. For any sets $A$ and $X$:

$A = A \cap (A \cup X)$ [By Absorption Law of Sets]

Similarly, for sets $B$ and $X$:

$B = B \cap (B \cup X)$ [By Absorption Law of Sets]

Step 2: Applying the Equality Constraints

We are given the condition that $A \cup X = B \cup X$. Let this union be represented by a common set $U_x$. Thus:

$A = A \cap U_x$

$B = B \cap U_x$

Step 3: Utilizing the Distributive Law

Since $U_x = A \cup X$, we substitute this back into the expression for $A$:

$A = A \cap (A \cup X)$

Using the Distributive Law of Intersection over Union, $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$, we expand the expression for the intersection of $A$ with $(B \cup X)$:

$A = A \cap (B \cup X) = (A \cap B) \cup (A \cap X)$ [Distributive Law]

Step 4: Applying Null Set Properties

It is given that $A \cap X = \phi$. Substituting this into our equation:

$A = (A \cap B) \cup \phi$

$A = A \cap B$ [Identity Law: $S \cup \phi = S$]

Step 5: Symmetry of the Argument

We perform the identical derivation for set $B$:

$B = B \cap (A \cup X)$ [Since $A \cup X = B \cup X$]

$B = (B \cap A) \cup (B \cap X)$ [Distributive Law]

Given $B \cap X = \phi$, we have:

$B = (B \cap A) \cup \phi$

$B = B \cap A$ [Identity Law]

Since the intersection operation is commutative ($A \cap B = B \cap A$), it follows that $A = B$.

Final Solution: $A = B$ is proven via the application of the Distributive and Absorption laws, constrained by the provided intersectional nullity and union equality.

Step 1: Theoretical Premise & Counter-Example Strategy

To demonstrate that $A \cap B = A \cap C$ does not universally imply $B = C$, we employ the method of disproof by counter-example. In set theory, the intersection operator is not injective; that is, the mapping $f(X) = A \cap X$ is not necessarily one-to-one for a fixed set $A$ unless $A$ is a universal set ($U$).

Step 2: Defining the Sets for Counter-Example

Let us construct sets $A, B,$ and $C$ such that the intersections are identical, yet the sets $B$ and $C$ remain distinct.

• Let $A = \{1, 2\}$
• Let $B = \{1, 3\}$
• Let $C = \{1, 4\}$

[Justification: By selecting elements outside of $A$ to be unique to $B$ and $C$, we isolate the intersection to the common element(s) shared with $A$.]

Step 3: Calculating $A \cap B$ and $A \cap C$

Applying the definition of the intersection of two sets ($A \cap X = \{x : x \in A \text{ and } x \in X\}$):

$A \cap B = \{1, 2\} \cap \{1, 3\} = \{1\}$

$A \cap C = \{1, 2\} \cap \{1, 4\} = \{1\}$

[Per the axiom of extensionality, since both intersections result in the singleton set $\{1\}$, we confirm $A \cap B = A \cap C$.]

Step 4: Logical Comparison and Conclusion

We observe the following:

Visual Representation of Set Relationships

The discrepancy arises because the elements $3 \in B$ and $4 \in C$ are not present in set $A$. Consequently, they are "ignored" by the intersection operation. Because the intersection operation eliminates elements that distinguish $B$ from $C$ (specifically elements in $B \setminus A$ and $C \setminus A$), the equality of the intersections provides insufficient information to determine the equality of the original sets $B$ and $C$.

Final Solution: Since $A \cap B = \{1\}$ and $A \cap C = \{1\}$ while $B \neq C$, we have shown that $A \cap B = A \cap C$ does not imply $B = C$.

Part 1: Proof of Equality $A = ( A \cap B ) \cup ( A - B )$

To prove that $A = ( A \cap B ) \cup ( A - B )$, we employ the method of double inclusion, demonstrating that the left-hand side (LHS) is a subset of the right-hand side (RHS) and vice versa.

Step 1.1: Proving $(A \cap B) \cup (A - B) \subseteq A$

• Let $x \in (A \cap B) \cup (A - B)$.
• By the definition of union, $x \in (A \cap B)$ or $x \in (A - B)$. [Definition of Union]
• If $x \in (A \cap B)$, then $x \in A$ and $x \in B$. Thus, $x \in A$. [Definition of Intersection]
• If $x \in (A - B)$, then $x \in A$ and $x \notin B$. Thus, $x \in A$. [Definition of Set Difference]
• In both cases, $x \in A$. Therefore, $(A \cap B) \cup (A - B) \subseteq A$.

Step 1.2: Proving $A \subseteq (A \cap B) \cup (A - B)$

• Let $x \in A$.
• Either $x \in B$ or $x \notin B$. [Law of Excluded Middle]
• Case 1: If $x \in B$, then since $x \in A$ and $x \in B$, it follows that $x \in (A \cap B)$.
• Case 2: If $x \notin B$, then since $x \in A$ and $x \notin B$, it follows that $x \in (A - B)$.
• Since $x$ must satisfy either Case 1 or Case 2, $x \in (A \cap B) \cup (A - B)$.
• Therefore, $A \subseteq (A \cap B) \cup (A - B)$.

Since both inclusions hold, $A = (A \cap B) \cup (A - B)$.

Part 2: Proof of Equality $A \cup (B - A) = A \cup B$

Step 2.1: Algebraic Manipulation using Distributive Law

We utilize the identity $B - A = B \cap A^c$, where $A^c$ is the complement of $A$.

$A \cup (B - A) = A \cup (B \cap A^c)$ [Definition of Relative Complement]

Applying the Distributive Law of Union over Intersection: $X \cup (Y \cap Z) = (X \cup Y) \cap (X \cup Z)$.

$A \cup (B \cap A^c) = (A \cup B) \cap (A \cup A^c)$ [Distributive Law]

Step 2.2: Applying Complement Laws

• We know that $A \cup A^c = U$, where $U$ is the Universal Set. [Complement Law]
• The expression becomes $(A \cup B) \cap U$.
• Any set intersected with the universal set returns the set itself: $X \cap U = X$. [Identity Law]
• $(A \cup B) \cap U = A \cup B$.

Final Solution: The identities $A = ( A \cap B ) \cup ( A - B )$ and $A \cup ( B - A ) = ( A \cup B )$ are proven via set inclusion and logical axioms respectively.