Study your flashcards anywhere!

Download the official Cram app for free >

  • Shuffle
    Toggle On
    Toggle Off
  • Alphabetize
    Toggle On
    Toggle Off
  • Front First
    Toggle On
    Toggle Off
  • Both Sides
    Toggle On
    Toggle Off
  • Read
    Toggle On
    Toggle Off
Reading...
Front

How to study your flashcards.

Right/Left arrow keys: Navigate between flashcards.right arrow keyleft arrow key

Up/Down arrow keys: Flip the card between the front and back.down keyup key

H key: Show hint (3rd side).h key

A key: Read text to speech.a key

image

Play button

image

Play button

image

Progress

1/105

Click to flip

105 Cards in this Set

  • Front
  • Back
Set Identities
A⋃∅
A⋂U
Identity Laws
A
A
Set Identities
A⋃U
A⋂∅
Domination Laws
U
Set Identities
A⋃A
A⋂A
Idempotent Laws
A
A
Set Identities
A⋃B
A⋂B
Commutative Laws
B⋃A
B⋂A
Set Identities
A⋃(B⋃C)
A⋂(B⋂C)
Associative Laws
(A⋃B)⋃C
(A⋂B)⋂C
Set Identities
A⋂(B⋃C)
A⋃(B⋂C)
Distributive Laws
(A⋂B)⋃(A⋂C)
(A⋃B)⋂(A⋃C)
Set Identities
(A⋂B)⋃(A⋂C)
(A⋃B)⋂(A⋃C)
Reverse Distributive Laws
A⋂(B⋃C)
A⋃(B⋂C)
Set Identities
~(A⋃B)
~(A⋂B)
Those should be lines over the whole top
De Morgan's Laws
~A⋂~B
~A⋃~B
Those should be lines over individual letters
Set Identities
~A⋂~B
~A⋃~B
Those should be lines over individual letters
Reverse De Morgan's Laws
~(A⋃B)
~(A⋂B)
Those should be lines over the whole top
Set Identities
A⋃(A⋂B)
A⋂(A⋃B)
Absorption Laws
A
A
Set Identities
A⋃~A
A⋂~A
Should be lines on top
Complement Laws
U
Truth sets
{x|P(x)}

explain the parts
The elements of the set for which the values of x make the second part true - the elementhood test
y∈{x|P(x)}
P(y)
y∉{x|P(x)}
~P(y)
A is a truthset for the proposition P(x)
What is A in set form
What is meant by
y∈A
y∉A
P(y)
~P(y)
Redefine
{x∈U|P(x)}
{x|x∈U^P(x)}
Redefine
{x|x∈U^P(x)}
{x∈U|P(x)}
Redefine
y∈{x∈U|P(x)}
y∈U^P(y)
Redefine: What is it?
A⋂B
It is a set
{x|x∈A^x∈B}

If A and B have truth tests, then it could mean P(x)^Q(x)
What is it?
x ∈ A⋂B
It is a statement
x∈A^x∈B

and if they had propositions
P(x)^Q(x)
Redefine: What is it?
A⋃B
It is a set
{x|x∈A v x∈B}

If A and B have truth tests, then it could mean P(x) v Q(x)
What is it?
x ∈ A⋃B
It is a statement
x∈A v x∈B

and if they had propositions
P(x) v Q(x)
Redefine: What is it?
A/B
It is a set
{x|x∈A ^ x∉B}

If A and B have truth tests, then it could mean P(x) ^ ~Q(x)
What is it?
x ∈ A/B
It is a statement
x∈A ^ x∉B

and if they had propositions
P(x) ^ ~Q(x)
Redefine:
{x|x∈A^x∈B}
A⋂B
Redefine:
x∈A^x∈B
x ∈ A⋂B
Redefine:
{x|x∈A v x∈B}
A⋃B
x∈A v x∈B
x ∈ A⋃B
Redefine:
{x|x∈A ^ x∉B}
A/B
Redefine:
x∈A ^ x∉B
x ∈ A/B
What is it? Name it
Redefine:
A∆B

(2 definitions)
It's a set
Symmetric Difference
(A\B)⋃(B\A)
(A⋃B)\(A⋂B)
Redefine: what is it?

A⊆B
(2 definitions)
It is a statement:

∀x(x∈A → x∈B)
∀x∈A(x∈B)
Redefine: What is it?

A=B
(2 definitions)
It is a Statement

∀x(x∈A ↔ x∈B)
A⊆B ^ B⊆A
Redefine:
(A\B)⋃(B\A)
(A⋃B)\(A⋂B)
A∆B
Redefine:
∀x(x∈A → x∈B)
∀x∈A(x∈B)
A⊆B
Redefine:
∀x(x∈A ↔ x∈B)
A⊆B ^ B⊆A
A=B
What is it? What is it called?
Redefine
A⋂B=∅
A statement
Called disjoint
Allx, x is part of the first part iff x is part of the second part?
or This first part is a subset of the second, and the second is a subset of the first?
What is it?
Re-express:

{pi|i∈I}
An indexed set, the set contains all numbers pi with i being the element of some set.
{x|∃i∈I(x=pi)}
What is it?
Re-express:

{x|∃i∈I(x=pi)}
An indexed set, the set contains all numbers pi with i being the element of some set.

{pi|i∈I}
An indexed set, the set contains all numbers pi with i being the element of some set.

Give two ways
{x|∃i∈I(x=pi)}

{pi|i∈I}
Re-express the family of sets F in bracket notation:
using the classes of students Cs

give two ways
Each Cs is a set itself
{{1 2},{3,4},{5,6}} C1 = {1,2}

F = {Cs|s∈S}
{X|∃s∈S(X=Cs)}
What is it?
Re-express:

F = {Cs|s∈S}
Indexed Family
Each Cs is a set itself
{{1 2},{3,4},{5,6}} C1 = {1,2}

{X|∃s∈S(X=Cs)}
What is it?
Re-express:

F = {X|∃s∈S(X=Cs)}
Indexed Family
Each Cs is a set itself
{{1 2},{3,4},{5,6}} C1 = {1,2}

F = {Cs|s∈S}
What is it?
Re-express

P(A)
The power set:

{x|x⊆A}

x is a set
What is it?
Re-express

{x|x⊆A}

x is a set
The power set:
P(A)
The family set is a subset of what?
Give reasoning
If all courses were set C then each Cs is a subset of C

For each student Cs∈P(C)
Every element then of the family is an element of P(c) so F ⊆ P(C)
Equivalence

P(A⋂B)
P(A)⋂P(B)
Equivalence

P(A)⋂P(B)
P(A⋂B)
Explain and define:

⋂F
F = {{1,2},{2,3},{3,4}}
this is the intersection of the family
{1,2}⋂{2,3}⋂{3,4}
{x|∀A∈F(x∈A)}
{x|∀A(A∈F → x∈A)}
What is this?/

{x|∀A∈F(x∈A)}
F = {{1,2},{2,3},{3,4}}
this is the intersection of the family
{1,2}⋂{2,3}⋂{3,4}
{x|∀A∈F(x∈A)}
{x|∀A(A∈F → x∈A)}
What is this?

{x|∀A(A∈F → x∈A)}
F = {{1,2},{2,3},{3,4}}
this is the intersection of the family
{1,2}⋂{2,3}⋂{3,4}
{x|∀A∈F(x∈A)}
{x|∀A(A∈F → x∈A)}
What is this?

{1,2}⋂{2,3}⋂{3,4}
F = {{1,2},{2,3},{3,4}}
this is the intersection of the family
{1,2}⋂{2,3}⋂{3,4}
{x|∀A∈F(x∈A)}
{x|∀A(A∈F → x∈A)}
Explain and define:

⋃F
F = {{1,2},{2,3},{3,4}}
this is the union of the family
{1,2}⋃{2,3}⋃{3,4}
{x|∃A∈F(x∈A)}
{x|∃A(A∈F ^ x∈A)}
What is this?/

{x|∃A∈F(x∈A)}
F = {{1,2},{2,3},{3,4}}
this is the union of the family
{1,2}⋃{2,3}⋃{3,4}
{x|∃A∈F(x∈A)}
{x|∃A(A∈F ^ x∈A)}
What is this?

{x|∃A(A∈F ^ x∈A)}
F = {{1,2},{2,3},{3,4}}
this is the union of the family
{1,2}⋃{2,3}⋃{3,4}
{x|∃A∈F(x∈A)}
{x|∃A(A∈F ^ x∈A)}
What is this?

{1,2}⋃{2,3}⋃{3,4}
F = {{1,2},{2,3},{3,4}}
this is the union of the family
{1,2}⋃{2,3}⋃{3,4}
{x|∃A∈F(x∈A)}
{x|∃A(A∈F ^ x∈A)}
Define

⋂F=∅
undefined
What is it?
Redefine
⋂(i∈I) Ai
Where F puts Ai as sets in a set, this is more like the union of all the sets, it is the same thing as ⋂F
{x|∀i∈I(x∈Ai)}
{x|∀i(i∈I → x∈Ai)}
It asks that all the sets have the same element
What is it?

{x|∀i∈I(x∈Ai)}
⋂(i∈I) Ai
Where F puts Ai as sets in a set, this is more like the union of all the
sets, it is the same thing as ⋂F
{x|∀i∈I(x∈Ai)}
{x|∀i(i∈I → x∈Ai)}
It asks that all the sets have the same element
What is it?
Redefine
⋃(i∈I) Ai
Where F puts Ai as sets in a set, this is more like the union of all the sets, it is the same thing as ⋃F
{x|∃i∈I(x∈Ai)}
{x|∃i(i∈I ^ x∈Ai)}
That at least one of the sets of Ai have the element
What is it?
Redefine

{x|∃i∈I(x∈Ai)}
Where F puts Ai as sets in a set, this is more like the union of all the
⋃(i∈I) Ai
How is this different from a indexed set or indexed family? Because it's an element, not equal
sets, it is the same thing as ⋃F
{x|∃i∈I(x∈Ai)}
{x|∃i(i∈I ^ x∈Ai)}
That at least one of the sets of Ai have the element
What is it?
Define

AxB
Cartesian Product

{(a,b)| a∈A ^ b∈B}

results in all ordered pairs
What is it?
Define

{(a,b)| a∈A ^ b∈B}
Cartesian Product
AxB

results in all ordered pairs
Equivalence:
Ax(B⋂C)
(AxB)⋂(AxC)
Equivalence:
(AxB)⋂(AxC)
Ax(B⋂C)
Equivalence:
Ax(B⋃C)
(AxB)⋃(AxC)
Equivalence:
(AxB)⋃(AxC)
Ax(B⋃C)
Equivalence:
(AxB)⋂(CxD)
(A⋂C)x(B⋂D)
Equivalence:
(A⋂C)x(B⋂D)
(AxB)⋂(CxD)
Subset:

(AxB)⋃(CxD) ⊆
(A⋃C)x(B⋃D)
Equivalence:

Ax∅
∅xA
AxB = BxA if and only if
A =∅ or
B = ∅ or
A=B
Truth set of P(x,y)

two versions
{(a,b)∈AxB | P(x,y)}
{(a,b) | (a,b)∈AxB ^ P(x,y)}
{(a,b)∈AxB | P(x,y)}
Truth set of P(x,y)
{(a,b) | (a,b)∈AxB ^ P(x,y)}
Truth set of P(x,y)
Define a relation from A to B
R ⊆ AxB
does not need to imply a truth set for R, it can be any subset
What is this?
R ⊆ AxB
A relation
Domain of R for AxB
{a∈A | ∃b∈B((a,b)∈R}
{a∈A | ∃b∈B((a,b)∈R}
Domain of R
Range of R
{b∈B | ∃a∈A((a,b)∈R}
{b∈B | ∃a∈A((a,b)∈R}
Range of R
Inverse of R R-1
{(b,a)∈BxA | (a,b)∈R}
{(b,a)∈BxA | (a,b)∈R}
Inverse of R R-1
Composition of two relations
R⊆AxB
S⊆BxC
SoR
{(a,c)∈AxC |
∃b∈B((a,b)∈R ^ (b,c)∈S}
{(a,c)∈AxC |
∃b∈B((a,b)∈R ^ (b,c)∈S}
Composition of two relations
R⊆AxB
S⊆BxC
SoR
(r,s)∈L-1 if and only if
(s,r)∈L
(s,p)∈ToE if and only if
∃c( (s,c)∈E ^ (c,p) ∈ T
(x,y) ∈ R
xRy
xRy
(x,y) ∈ R
Relation equivalence:
(R-1)-1
R
Relation equivalence:
Relation equivalence:Dom(R-1)
Ran(R)
Relation equivalence:
Ran(R)
Dom(R-1)
Relation equivalence:
Ran(R-1)
Dom(R)
Relation equivalence:
Dom(R)
Ran(R-1)
Relation equivalence:
To(SoR)
(ToS)oR
Relation equivalence:
(ToS)oR
To(SoR)
(SoR)-1
R-1 o S-1
Equivalence Relation:
R-1 o S-1
(SoR)-1
the relation R is reflexive on A (or just reflexive if A is clear from context) if

(2 versions)
∀x∈A(xRx)
or
∀x∈A((x,x)∈Rx)
the relation R is symmetric if
∀x∈A∀y∈A(xRy → yRx)
or
∀x∈A∀y∈A((x,y)∈R → (y,x)∈R)
The relation R is transitive if
∀x∈A∀y∈A∀z∈A
((xRy ^ yRz) → xRz)

also can be done without shorthand
R is reflexive iff
i(A) ⊆ R, where as before i(A) is the identity relation on A
R is symmetric iff
R=R-1
R is transitive iff
RoR ⊆ R
∃!xP(x)

4 definitions
∃x(P(x)^~∃y(P(y) ^ y≠x))
∃x(P(x)^∀y(P(y)→y=x))
∃xP(x) ^ ∀y∀z[(P(y)^P(z))→y=z]
Existence Uniqueness
∃x∀y(P(y)↔y=x)