nyuuzyou/wikis huggingface dataset:
Set Difference and Intersection form Partition/Corollary 1

== Corollary to Set Difference and Intersection form Partition ==
Let $S$ and $T$ be sets such that:
:$S \setminus T \ne \O$
:$T \setminus S \ne \O$
:$S \cap T \ne \O$
Then $S \setminus T$, $T \setminus S$ and $S \cap T$ form a partition of $S \cup T$, the union of $S$ and $T$.
== Proof ==
From Set Difference and Intersection form Partition:
:$S \setminus T$ and $S \cap T$ form a partition of $S$
:$T \setminus S$ and $S \cap T$ form a partition of $T$
From Set Difference is Disjoint with Reverse:
:$\paren {S \setminus T} \cap \paren {T \setminus S} = \O$
So:
:$S \cup T = \paren {S \setminus T} \cup \paren {S \cap T} \cup \paren {T \setminus S} \cup \paren {S \cap T}$
and the result follows.
Category:Set Difference
Category:Set Intersection
Category:Set Union
Category:Set Difference and Intersection form Partition
Set Difference and Intersection form Partition/Corollary 2

== Corollary to Set Difference and Intersection form Partition ==
Let $\O \subsetneqq T \subsetneqq S$.
Then:
:$\set {T, \relcomp S T}$
is a partition of $S$.
== Proof ==
First we note that:
From the definition of relative complement, we have:
:$\relcomp S T = S \setminus T$.
It follows from Set Difference with Self is Empty Set that:
:$T \ne S \iff \relcomp S T \ne \O$
From Intersection with Relative Complement is Empty:
:$T \cap \relcomp S T = \O$
That is, $T$ and $\relcomp S T$ are disjoint.
From Union with Relative Complement:
:$T \cup \relcomp S T = S$
That is, the union of $T$ and $\relcomp S T$ forms the whole set $S$.
Thus all the conditions for a partition are satisfied.
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Example $6.4$
Category:Set Partitions
Category:Relative Complement
Category:Set Difference
Category:Set Difference and Intersection form Partition
Set Difference as Intersection with Complement

== Theorem ==
Set difference can be expressed as the intersection with the set complement:
:$A \setminus B = A \cap \map \complement B$
== Proof ==
This follows directly from Set Difference as Intersection with Relative Complement:
:$A \setminus B = A \cap \relcomp S B$
Let $S = \Bbb U$.
Since $A, B \subseteq \Bbb U$ by definition of the universal set, the result follows.
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B x}$
* : $\S 2$: Problem $1 \ \text{(i)}$
Category:Set Complement
Category:Set Difference
Category:Set Intersection
Set Difference as Intersection with Relative Complement

== Theorem ==
Let $A, B \subseteq S$.
Then the set difference between $A$ and $B$ can be expressed as the intersection with the relative complement with respect to $S$:
:$A \setminus B = A \cap \relcomp S B$
== Proof ==
== Sources ==
* : $\S 5$: Complements and Powers
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
* : Chapter $1$: Sets and Logic: Exercise $5$
* : $1$: Events and probabilities: $1.2$: Outcomes and events: Exercise $2$
Category:Set Difference
Category:Relative Complement
Category:Set Intersection
Set Difference as Symmetric Difference with Intersection

== Theorem ==
:$S \setminus T = S \symdif \paren {S \cap T}$
where:
:$S \setminus T$ denotes set difference
:$S \symdif T$ denotes set symmetric difference
:$S \cap T$ denotes set intersection.
== Proof ==
Category:Set Intersection
Category:Set Difference
Category:Symmetric Difference
Set Difference is Anticommutative

== Theorem ==
Set difference is an anticommutative operation:
:$S = T \iff S \setminus T = T \setminus S = \O$
== Proof ==
From Set Difference with Superset is Empty Set‎ we have:
:$S \subseteq T \iff S \setminus T = \O$
:$T \subseteq S \iff T \setminus S = \O$
The result follows from definition of set equality:
:$S = T \iff \paren {S \subseteq T} \land \paren {T \subseteq S}$
== Also see ==
* Union is Commutative
* Intersection is Commutative
* Symmetric Difference is Commutative
== Sources ==
* : $\S 1.1$: Theorem $1.7$
* : Exercise $1.1: \ 7$
Category:Set Difference
Category:Empty Set
Set Difference is Disjoint with Reverse

== Theorem ==
:$\paren {S \setminus T} \cap \paren {T \setminus S} = \O$
== Proof ==
We assume that $S, T \subseteq \mathbb U$ where $\mathbb U$ is the universal set.
Then we can use the definition of Set Difference as Intersection with Complement.
Category:Set Difference
Category:Set Intersection
Set Difference is Right Distributive over Set Intersection

== Theorem ==
Let $A, B, C$ be sets.
Then:
:$\paren {A \cap B} \setminus C = \paren {A \setminus C} \cap \paren {B \setminus C}$
where:
:$A \cap B$ denotes set intersection
:$A \setminus C$ denotes set difference.
That is, set difference is right distributive over set intersection.
=== Illustration by Venn Diagram ===
=== General Case ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.4 \ \text{(b)}$
Category:Set Difference
Category:Set Intersection
Category:Set Difference is Right Distributive over Set Intersection
Set Difference is Right Distributive over Set Intersection/General Case

== Theorem ==
Let $U$ be a collection of sets.
Let $T$ be a set.
Then:
:$\ds \bigcap_{X \mathop \in U} \paren {X \setminus T} = \paren {\bigcap_{X \mathop \in U} X} \setminus T$
That is, the difference with an intersection equals the intersection of the differences.
== Proof ==
Category:Set Difference is Right Distributive over Set Intersection
Set Difference is Right Distributive over Set Intersection/Venn Diagram

== Theorem ==
== Proof ==
Demonstration by Venn diagram:
: 
Consider the diagram on the .
The red and yellow areas together form $A \cap B$.
The red area without the yellow area forms $\paren {A \cap B} \setminus C$.
Consider the diagram on the .
The red and orange areas together form $A \setminus C$.
The yellow and orange areas together form $B \setminus C$.
Their intersection is the orange area, which forms $\paren {A \setminus C} \cap \paren {B \setminus C}$.
It is seen that the red area on the is the same as the orange area on the .
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.4 \ \text{(b)}$
Category:Set Difference is Right Distributive over Set Intersection
Category:Examples of Venn Diagrams
Set Difference is Right Distributive over Union

== Theorem ==
Set difference is right distributive over union.
Let $R, S, T$ be sets.
Then:
:$\paren {R \cup S} \setminus T = \paren {R \setminus T} \cup \paren {S \setminus T}$
where:
:$R \setminus S$ denotes set difference
:$R \cup T$ denotes set union.
== Proof ==
Category:Set Difference
Category:Set Union
Category:Examples of Distributive Operations
Set Difference is Set

== Theorem ==
Let $x$ be a small class.
Let $A$ be a class.
Let $\map \MM B$ denote that $B$ is small.
Then: 
:$\map \MM {x \setminus A}$
== Proof ==
By Set Difference as Intersection with Relative Complement:
:$\paren {x \setminus A} = \paren {x \cap \map \complement A}$
Next, by Axiom of Subsets Equivalents, $\paren {x \cap \map \complement A}$ is small.
Therefore:
:$\map \MM {x \setminus A}$
== Sources ==
* : $\S 5.15$
Category:Class Theory
Set Difference is Subset

== Theorem ==
Set difference is a subset of the first set:
:$S \setminus T \subseteq S$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 8$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets
Category:Set Difference
Category:Subsets
Category:Set Difference is Subset
Set Difference is Subset of Union of Differences

== Theorem ==
Let $R, S, T$ be sets.
Then:
:$R \setminus S \subseteq \paren {R \setminus T} \cup \paren {T \setminus S}$
where:
:$S \subseteq T$ denotes subset
:$S \setminus T$ denotes set difference
:$S \cup T$ denotes set union.
== Proof ==
Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.
Category:Set Difference
Category:Set Union
Set Difference is not Associative

== Theorem ==
Let $R, S, T$ be sets.
The expression:
:$\paren {R \setminus S} \setminus T = R \setminus \paren {S \setminus T}$
holds exactly when $R \cap T = \O$.
Here $R \setminus S$ denotes set difference.
Thus, set difference is not associative.
== Proof ==
We assume a universal set $\mathbb U$ such that $R, S, T \subseteq \mathbb U$.
We have the identity Set Difference as Intersection with Complement:
:$R \setminus S = R \cap \overline S$
where $\overline S$ is the set complement of $S$:
:$\overline S = \relcomp {\Bbb U} S$
Thus we can represent the two expressions as follows:
For the second:
As can be seen, this full expansion of the second expression can be expressed:
:$R \setminus \paren {S \setminus T} = \paren {\paren {R \setminus S} \setminus T} \cup \paren {R \cap T}$
It directly follows from Intersection with Empty Set that:
:$R \setminus \paren {S \setminus T} = \paren {R \setminus S} \setminus T \iff R \cap T = \O$
== Also see ==
* Intersection is Associative
* Union is Associative
* Symmetric Difference is Associative
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $15$
Category:Set Difference
Set Difference of Cartesian Products

== Theorem ==
:$\paren {S_1 \times S_2} \setminus \paren {T_1 \times T_2} = \paren {S_1 \times \paren {S_2 \setminus T_2} } \cup \paren {\paren {S_1 \setminus T_1} \times S_2}$
== Proof ==
Let $\tuple {x, y} \in \paren {S_1 \times S_2} \setminus \paren {T_1 \times T_2}$.
Then:
The result follows from the definition of subset and set equality.
== Sources ==
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 9 \alpha$
Category:Cartesian Product
Category:Set Difference
Set Difference of Events is Event

== Theorem ==
Let $\EE$ be an experiment with a probability space $\struct {\Omega, \Sigma, \Pr}$.
The event space $\Sigma$ of $\EE$ has the property that:
:$A, B \in \Sigma \implies A \setminus B \in \Sigma$
That is, the difference of two events is also an event in the event space.
== Proof ==
== Also see ==
* Elementary Properties of Event Space
== Sources ==
* : $1$: Events and probabilities: $1.2$: Outcomes and events: Exercise $2$
* : $1$: Events and probabilities: $1.4$: Probability spaces: $(12)$
Category:Event Spaces
Set Difference of Intersection with Set is Empty Set

== Theorem ==
The set difference of the intersection of two sets with one of those sets is the empty set.
Let $S, T$ be sets.
Then:
:$\paren {S \cap T} \setminus S = \O$
:$\paren {S \cap T} \setminus T = \O$
== Proof ==
From Set Difference is Right Distributive over Set Intersection:
: $\paren {R \cap S} \setminus T = \paren {R \setminus T} \cap \paren {S \setminus T}$
Hence:
Category:Set Difference
Category:Set Intersection
Category:Empty Set
Set Difference of Larger Set with Smaller is Not Empty

== Theorem ==
Let $S$ and $T$ be finite sets.
Let $\card S > \card T$.
Then:
:$S \setminus T \ne \O$
== Proof ==
From Cardinality of Subset of Finite Set:
:$S \nsubseteq T$
From the contrapositive statement of Set Difference with Superset is Empty Set:
:$S \setminus T \ne \O$.
Category:Set Difference
Category:Cardinality
Set Difference of Matroid Dependent Set with Independent Set is Non-empty

== Theorem == 
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $I$ be an independent subset of $M$.
Let $D$ be a dependent subset of $M$.
Then:
:$D \setminus I \ne \O$
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof ==
From Independent Subset Contains No Dependent Subset:
:$D \nsubseteq I$
By definition of subset:
:$\exists x \in D : x \notin I$
By definition of set difference:
:$\exists x \in D \setminus I$
The result follows.
Category:Matroid Dependent Subsets
Category:Matroid Independent Subsets
Category:Set Difference of Matroid Dependent Set with Independent Set is Non-empty
Set Difference of Matroid Dependent Set with Independent Set is Non-empty/Corollary 1

== Theorem == 
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $D$ be a dependent subset of $M$.
Let $B$ be a base of $M$.
Then:
:$D \setminus B \ne \O$
== Proof ==
By definition of matroid base:
:$B$ is an independent subset of $M$
From Set Difference of Matroid Dependent Set with Independent Set is Non-empty:
:$D \setminus B \ne \O$
Category:Set Difference of Matroid Dependent Set with Independent Set is Non-empty
Set Difference of Matroid Dependent Set with Independent Set is Non-empty/Corollary 2

== Theorem == 
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $I$ be an independent subset of $M$.
Let $C$ be a circuit of $M$.
Then:
:$C \setminus I \ne \O$
== Proof ==
By definition of matroid circuit:
:$C$ is a dependent subset of $M$
From Set Difference of Matroid Dependent Set with Independent Set is Non-empty:
:$C \setminus I \ne \O$
Category:Set Difference of Matroid Dependent Set with Independent Set is Non-empty
Set Difference of Matroid Dependent Set with Independent Set is Non-empty/Corollary 3

== Theorem == 
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $B$ be an base of $M$.
Let $C$ be a circuit of $M$.
Then:
:$C \setminus B \ne \O$
== Proof ==
By definition of matroid base:
:$B$ is an independent subset of $M$
By definition of matroid circuit:
:$C$ is a dependent subset of $M$
From Set Difference of Matroid Dependent Set with Independent Set is Non-empty:
:$C \setminus B \ne \O$
Category:Set Difference of Matroid Dependent Set with Independent Set is Non-empty
Set Difference of Relations Compatible with Group Operation is Compatible

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\RR, \QQ$ be relations on $G$ which are compatible with $\circ$.
Then the difference $\RR \setminus \QQ$ is compatible with $\circ$.
== Proof ==
By Complement of Relation Compatible with Group is Compatible, $\relcomp {G \times G} \QQ$ is compatible with $\circ$.
Thus by Intersection of Relations Compatible with Operation is Compatible, $\RR \cap \relcomp {G \times G} \QQ$ is compatible with $\circ$.
But:
:$\RR \cap \relcomp {G \times G} \QQ = \RR \setminus \QQ$
so the theorem holds.
Category:Compatible Relations
Set Difference with Disjoint Set

== Theorem ==
Let $S, T$ be sets.
Then:
:$S \cap T = \O \iff S \setminus T = S$
where:
:$S \cap T$ denotes set intersection
:$\O$ denotes the empty set
:$S \setminus T$ denotes set difference.
== Proof ==
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : Appendix $\text{A}.2$: Theorem $\text{A}.11$
Category:Set Difference
Category:Disjoint Sets
Set Difference with Empty Set is Self

== Theorem ==
The set difference between a set and the empty set is the set itself:
:$S \setminus \O = S$
== Proof ==
From Set Difference is Subset:
:$S \setminus \O \subseteq S$
From the definition of the empty set:
:$\forall x \in S: x \notin \O$
Let $x \in S$.
Thus:
Thus we have:
:$S \setminus \O \subseteq S$
and:
:$S \subseteq S \setminus \O$
So by definition of set equality:
:$S \setminus \O = S$
== Also see ==
* Set Difference with Self is Empty Set
== Sources ==
* : Exercise $1.1: \ 8 \ \text{(b)}$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 8 \ \text{(d)}$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.5 \ \text{(i)}$
* : Appendix $\text{A}.2$: Theorem $\text{A}.11$
Category:Set Difference
Category:Empty Set
Set Difference with Intersection

== Theorem ==
Let $S$ and $T$ be sets.
=== Set Difference with Intersection is Difference ===
=== Set Difference of Intersection with Set is Empty Set ===
=== Set Difference Intersection with First Set is Set Difference ===
=== Set Difference Intersection with Second Set is Empty Set ===
=== Set Difference with Set Difference ===
== Also see ==
* De Morgan's Laws: Difference with Intersection
Category:Set Difference
Category:Set Intersection
Category:Empty Set
Set Difference with Intersection is Difference

== Theorem ==
The set difference with the intersection is just the set difference.
Let $S, T$ be sets.
Then:
:$S \setminus \paren {S \cap T} = S \setminus T$
== Proof ==
== Sources ==
* : $\S 1.1$: Theorem $1.6$
* : $\S 1.6$. Difference and complement: Example $20$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.3$
Category:Set Difference
Category:Set Intersection
Set Difference with Non-Empty Proper Subset is Non-Empty Proper Subset

== Theorem ==
Let $S$ be a set.
Let $T \subsetneq S$ be a non-empty proper subset of $S$.
 
Let $S \setminus T$ denote the set difference between $S$ and $T$.
Then:
:$S \setminus T$ is a non-empty proper subset of $S$
== Proof ==
From Set Difference is Subset:
:$S \setminus T \subseteq S$
From Set Difference with Proper Subset:
:$S \setminus T \ne \O$
Then :
:$T \ne \O$
From Intersection with Subset is Subset:
:$S \cap T = T$
Hence:
:$S \cap T \ne \O$
From the contrapositive statement of Set Difference with Disjoint Set:
:$S \setminus T \ne S$
It follows that $S \setminus T$ is a non-empty proper subset by definition.
Category:Set Difference
Category:Proper Subsets
Set Difference with Proper Subset

== Theorem ==
Let $S$ be a set.
Let $T \subsetneq S$ be a proper subset of $S$.
 
Let $S \setminus T$ denote the set difference between $S$ and $T$.
Then:
:$S \setminus T \ne \O$
where $\O$ denotes the empty set.
== Proof ==
 $S \setminus T = \O$.
Then:
:$\not \exists x \in S: x \notin T$
By De Morgan's laws:
:$\forall x \in S: x \in T$
By definition of subset:
:$S \subseteq T$
By definition of proper subset, we have that $T \subseteq S$ such that $T \ne S$.
But we have $T \subseteq S$ and $S \subseteq T$.
So by definition of set equality:
:$S = T$
From this contradiction it follows that:
:$S \setminus T \ne \O$
Category:Set Difference
Category:Proper Subsets
Set Difference with Self is Empty Set

== Theorem ==
The set difference of a set with itself is the empty set:
:$S \setminus S = \O$
== Proof ==
From Set is Subset of Itself:
:$S \subseteq S$
From Set Difference with Superset is Empty Set‎ we have:
:$S \subseteq T \iff S \setminus T = \O$
Hence the result.
== Also see ==
* Set Difference with Empty Set is Self
== Sources ==
* : Exercise $1.1: \ 8 \ \text{(a)}$
* : $\S 1.6$. Difference and complement: : Example $19$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 8 \ \text{(d)}$
* : $\S 5.17$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.5 \ \text{(ii)}$
Category:Set Difference
Category:Empty Set
Set Difference with Set Difference

== Theorem ==
The set difference with the set difference of two sets is the intersection of the two sets:
:$S \setminus \paren {S \setminus T} = S \cap T = T \setminus \paren {T \setminus S}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $2 \ \text{(a)}$
* : $\S 5$: Complements and Powers
* : A Note on Set-Theoretic Concepts: $(5)$
* : Exercise $1.1: \ 6$
* : Chapter $1$: Sets and Logic: Exercise $5$
* : $\S 2$: Problem $1 \ \text{(iii)}$
Category:Set Difference with Set Difference
Category:Set Intersection
Category:Set Difference
Set Difference with Set Difference/Proof 1

== Theorem ==
== Proof ==
Interchanging $S$ and $T$:
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $2 \ \text{(a)}$
* : $\S 5$: Complements and Powers
* : A Note on Set-Theoretic Concepts: $(5)$
* : Exercise $1.1: \ 6$
* : Chapter $1$: Sets and Logic: Exercise $5$
* : $\S 2$: Problem $1 \ \text{(iii)}$
Category:Set Difference with Set Difference
Set Difference with Set Difference/Proof 2

== Theorem ==
== Proof ==
From the , all sets are classes.
The result then follows from Class Difference with Class Difference.
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $2 \ \text{(a)}$
* : $\S 5$: Complements and Powers
* : A Note on Set-Theoretic Concepts: $(5)$
* : Exercise $1.1: \ 6$
* : Chapter $1$: Sets and Logic: Exercise $5$
* : $\S 2$: Problem $1 \ \text{(iii)}$
Category:Set Difference with Set Difference
Set Difference with Set Difference is Union of Set Difference with Intersection

== Theorem ==
Let $R, S, T$ be sets.
Then:
:$R \setminus \paren {S \setminus T} = \paren {R \setminus S} \cup \paren {R \cap T}$
where:
:$S \setminus T$ denotes set difference
:$S \cup T$ denotes set union
:$S \cap T$ denotes set intersection.
=== Illustration by Venn Diagram ===
=== Corollary ===
== Proof ==
Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.4 \ \text{(d)}$
Category:Set Difference
Category:Set Union
Category:Set Intersection
Category:Set Difference with Set Difference is Union of Set Difference with Intersection
Set Difference with Set Difference is Union of Set Difference with Intersection/Corollary

== Theorem ==
Let $S$ and $T$ be sets.
Then:
:$T \setminus \paren {S \setminus T} = T$
where $S \setminus T$ denotes set difference.
== Proof ==
From Set Difference with Set Difference is Union of Set Difference with Intersection:
:$R \setminus \paren {S \setminus T} = \paren {R \setminus S} \cup \paren {R \cap T}$
where $R, S, T$ are sets.
Hence:
Category:Set Difference
Category:Set Union
Category:Set Difference with Set Difference is Union of Set Difference with Intersection
Set Difference with Set Difference is Union of Set Difference with Intersection/Venn Diagram

== Theorem ==
== Proof ==
Demonstration by Venn diagram:
: 
Consider the diagram on the .
The yellow area forms $S \setminus T$.
The red area forms $R \setminus \paren {S \setminus T}$.
Consider the diagram on the .
The yellow and orange areas together form $R \cap T$.
The red and orange areas together form $R \setminus S$.
The red, orange and yellow areas together form $\paren {R \setminus S} \cup \paren {R \cap T}$.
It is seen that the red area on the is the same as the red, orange and yellow areas together on the .
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.4 \ \text{(d)}$
Category:Set Difference with Set Difference is Union of Set Difference with Intersection
Category:Examples of Venn Diagrams
Set Difference with Subset is Superset of Set Difference

== Theorem ==
Let $A, B, S$ be sets or classes.
Suppose that $A \subseteq B$.
Then:
:$S \setminus B \subseteq S \setminus A$
where $\setminus$ represents set difference.
== Proof ==
Let $x \in S \setminus B$.
Then by definition of set difference:
:$x \in S$ and $x \notin B$
 $x \in A$.
Then since $A$ is a subset (or subclass) of $B$, $x \in B$, a contradiction.
Thus $x \notin A$.
Since $x \in S$ and $x \notin A$, we conclude that $x \in S \setminus A$.
As this holds for all $x \in S \setminus B$:
:$S \setminus B \subseteq S \setminus A$
Category:Set Difference
Set Difference with Superset is Empty Set

== Theorem ==
:$S \subseteq T \iff S \setminus T = \O$
where:
:$S \subseteq T$ denotes that $S$ is a subset of $T$
:$S \setminus T$ denotes the set difference between $S$ and $T$
:$\O$ denotes the empty set.
== Proof ==
== Sources ==
* : $\S 5$: Complements and Powers
* : $\S 1.6$. Difference and complement: Example $19$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 8 \ \text{(a)}$
* : $\S 1.5$: Complementation
* : Appendix $\text{A}.2$: Boolean Operations: Problem $\text{A}.2.1$
* : Appendix $\text{A}.2$: Theorem $\text{A}.11$
Category:Set Difference
Category:Subsets
Category:Empty Set
Set Difference with Union

== Theorem ==
Let $R, S, T$ be sets.
Then:
:$R \setminus \paren {S \cup T} = \paren {R \cup T} \setminus \paren {S \cup T} = \paren {R \setminus S} \setminus T = \paren {R \setminus T} \setminus S$
where:
:$R \setminus S$ denotes set difference
:$R \cup T$ denotes set union.
=== Illustration by Venn Diagram ===
== Proof ==
Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.
Then:
Then:
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.4 \ \text{(c)}$
* : $\S 2$: Problem $1 \ \text{(ii)}$
Category:Set Difference
Category:Set Union
Category:Set Difference with Union