Basic non-obvious things to remember

We can only write $a_1 + \ldots + a_n$, without parenthesis, because of the associative property of the real numbers.

(Definition) Identity: given a set $S$ and an operation $\square$ on $S$, it's the element $e$ for which
$$\forall x \in S, x \square e = x$$

(Definition) Inverse:  given an operation $\square$ on set $S$, we say $\square$ has an inverse operation $^-$ if
$$\forall x \in S, x \square x^- = e$$

Commutative property is not a general rule: $$a + b = b + a, a - b \neq b - a$$

Division by 0 is undefined because division is defined in terms of multiplication: $\frac a b = a \cdot b^{-1}$, and $0^{-1}$ is undefined.

Given that  $a \cdot b = a \cdot c$, one could say that we can "cut" $a$, and say that $b = c$. Well, if $a = 0$, it's not true! But, when $a \neq 0$, it is, otherwise it's important to be aware of the fact that "cutting" $a$ is hiding this set of  equations:
\begin{array}{c} a \cdot b & = a \cdot c \\\\ (a^{-1})\cdot(a \cdot b) & = (a^{-1})\cdot(a \cdot c) \end{array}

Notes on Digital Image Processing - I

(Definition) Image: it's a two-dimentional function $f(x,y)$ where $x$ and $y$ are spatial cartesian coordinates and the intensity of $f$ at any $(x,y)$ is called the intensity or gray level of that image at that point.

(Definition) Digital Image: When $x,y,f$ are all finite, we call the image a digital image.

Digital Image Processing has two main general purposes: improvement of pictorial information for human interpretation, and image processing for storage, transmission and representation for autonomous machine perception.

It's not easy to clearly differentiate the area of digital image processing from the area of computer vision (which simulate the human vision, involving learning and taking actions based on input images), there's a continuum between them. But this can be divided into three intervals, represented by their processes: low-, mid- and high-level processes. The first one encompasses the image preprocessing: noise reduction, contrast enhancement and image sharpening, being both the input and output images. The second one is about segmentation (partitioning of the image into areas and objects), description of those segments in order to make them intelligible for computer processing,  and classification (recognition) of individual objects. The latter one is about assign meaning to combinations of recognized objects.

The overlap area between image processing and image analysis, based on the previous paragraph, is the recognition of indivial regions and objects in an digital image. So, a definition of digital image processing can be stated as:

(Definition) Digital Image Processing: set of processes whose input and output are images and also processes that extract attributes from images, up to and including the recognition of individual objects.

In order to check some of the various applications of this field, it's a good approach to classify images according to it's souce of energy. Below, some application from the eletromagnetic waves are listed:

• Gamma-ray Imaging:  nuclear medicine and astronomical observations;
• X-ray Imaging: medical diagnosis, industry, astronomy;
• Ultraviolet band: lithografy, industrial inspection, microscopy, lasers, biological imaging, astronomical observations;
• Visible and Infrared bands: light microscopy, astronomy, remote sensing, industry, law enforcement;
• Microwave band: radar;
• Radio Band: medicine and astronomy;

Besides the eletromagnetic waves, there are other important sources of energy that can form images, as seen below:

• Acoustic imaging: geological exploration, industry, medicine;
• Electron microscopy:
• Synthetic imaging: fractals, 3-D modeling;

In order to work with digital image processing, it's important to be aware of some fundamental steps, which can be grouped in two categories: the ones related to processes for which the input and the output are images, and the ones for which the output is made of attributes extracted from the images. These steps are:

• Image acquisition: sometimes involves acquiring the image and doing some preprocessing such as scaling;
• Image enhancement: process of manipulating an image so thar the result is more suitable than the original for a specific application;
• Image restoration: a very objective process based on mathematical or probabilistic models of image degradation;
• Color image processing
• Wavelets: foundation for representing images in various degrees of resolution;
• Compression: techniques for reducing the storage required to save an image, or the bandwidth required to transmi it;
• Morphological processing: deals with tools for extracting image components that are useful in the representation and description of shape;
• Segmentation: procedures to partitionate an image into its constituent parts or objects;
• Representation and description: almost always follows the output of the segmentation phase, which is usually raw pixel data; involves deciding if the data should be represented as boundaries or regions; the description (feature selection) deals with extracting attributes that result in some quantitative information of interest or are basic for differentiating one class of object from another;
• Recognition: assigns labels to an object based on its descriptors.

All of these stages are connected to (and by) a knowledge base, which is the knowledge about a problem domain. This base is generally coded into an image processing system in the form of a knowledge database.

It's worth to point that a digital image processing task may not require all of those processes, it all depents of it's complexity.

Digital Image Processing Systems have a set of componentes described below:

• Image sensors: two elements are required, the physical device to capture the energy irradiated by the object we wish to image, and a digitizer for converting the output of the physical sensing device into digital form;
• Specialized image processing hardware: the digitizer plus hardware that performs other primitive operations, such as an arithmetic logic unit (ALU);
• Computer: general purpose computer, can range from a PC to a supercomputer;
• Software: specialized modules that perform specific tasks;
• Mass Storage
• Image displays
• Hardcopy: laser printers, film cameras, heat-sensitive devices, inkjet units, digital units;
• Networking

General concepts about functions

Function

It's a mathematical object responsible for transforming objects from a set A to a set B, preserving their identities. We say
$$f : A \to B$$ to indicate a function that takes elements from a set A and transforms them into elements of a set B. Preserving the identity of the objects in A means that, if we take two elements of A and they are the same, a function from A to B must send them to an unique object in B. In mathematical chattering: $$\forall x_1,x_2 \in A, x_1 = x_2 \Rightarrow f(x_1) = f(x_2)$$

One-to-one functions

When a function is one-to-one, it preserves the difference of the objects in A when transforming into objects of B. It means that this kind of function guarantees the following property: $$\forall x_1,x_2 \in A, x_1 \neq x_2 \Rightarrow f(x_1) \neq (x_2)$$

Onto functions

An onto function is capable of producing the set B, because all of the $y \in B$ has a correspondence in A; it means that $$\forall y \in B, \exists x \in A, f(x) = y$$

Bijective functions

When a function is both one-to-one and onto, we say that it's bijective. In essence, when there exist such a function from a set A to B, we say that A and B are equivalent or that the set A can be rewritten as the set B, and vice versa. It's a perfect translation process. It's like writing a number from a base X to a base Y.

Important functions

Constant function

A function $f : A \to B$ is said to be constant if this property holds: $$\forall x \in A, f(x) = b, b \in B$$

Identity function

A function $Id_A : A \to A$ of a set A sends every element of A to itself. It means that $$\forall x \in A, Id_A(x) = x$$

Singleton

A singleton $f_b : \{a\} \to B$, with $\{a\}$ being any unit set, sends the element $a$ to an element $b \in B$. In other words, $f_b(a) = b \in B$. For a better understanding, think of a singleton as a function related to an element of B, with the purpose of pointing to $b$ no matter what it receives. Note that every element of $B$ has a an associated singleton.

Projection

Given a cartesian product $A \times B$, a projection extracts one element of the ordered pairs. So, in this case, there are two projections: $$\pi_1 : A \times B \to A, \pi_1( \langle a,b \rangle ) = a$$ $$\pi_2 : A \times B \to B, \pi_2( \langle a,b \rangle ) = b$$

The set of all total functions from A to B

It's denoted by $B^A$ or $[A \to B]$.

Métodos de prova

Generalização Universal (GU)

Seja $x$ um elemento arbitrário,
então $P(x)$.
$\forall x$, $P(x)$

Demonstração Direta (DD)

Suponha $p$.
Então $q$.
Logo, se $p$ então $q$.

Introdução da Conjunção (IC)

Afirmação 1.
Afirmação 2.
Então Afirmação 1 $\land$ Afirmação 2.

Generalização Existencial (GE)

Tome um $k \mid P(k)$.
Logo, $\exists x \mid P(x)$.

Redução ao Absurdo (RA)

Suponha $p$.
Absurdo.
Logo, $\neg p$.

Instanciação Existencial (IE)

Existe $x \mid P(x)$.
Então $P(z)$.

Eliminação 1 da Conjunção ($E_1C$)

Afirmação 1 $\land$ Afirmação 2.
Logo, Afirmação 1.

Eliminação 2 da Conjunção ($E_2C$)

Afirmação 1 $\land$ Afirmação 2.
Logo, Afirmação 2.

Introdução 1 da Disjunção ($I_1D$)

Afirmação 2.
Então Afirmação 1 $\lor$ Afirmação 2.

Introdução 2 da Disjunção ($I_2D$)

Afirmação 1.
Então Afirmação 1 $\lor$ Afirmação 2.

Eliminação do Condicional ou Modus Ponens(MP)

Se $p$ então $q$.
$p$.
Logo, $q$.

Modus Tollens* (MT)

Se $p$, então $q$.
$\neg q$.
Então $\neg p$.

Contra-positiva* (CP)

Se $p$ então $q$.
Portanto, se $\neg q$ então $\neg p$.

Lei 1a de De Morgan (DM1a)

$\neg(p \land q)$
Logo, $\neg p \lor \neg q$.

Lei 2a de De Morgan (DM2a)

$\neg p \lor \neg q$.
Logo, $\neg(p \land q)$

Lei 1b de De Morgan (DM1b)

$\neg(p \lor q)$
Logo, $\neg p \land \neg q$.

Lei 2b de De Morgan (DM2b)

$\neg p \land \neg q$.
Logo, $\neg(p \lor q)$

Silogismo Disjuntivo* (SD)

$p \lor q$.
$\neg p$.
Então $p$.

Silogismo Hipotético* (SH)

Se $p$ então $q$.
Se $q$ então $r$.
Logo, se $p$ então $r$.

Lei da Exportação* (LE)

Se $p$ então (se $q$ então $r$).
Então, se $p$ então $q$, então $r$.

Proving the uniqueness of the empty set

While proving some assertion about sets, the need for showing that a set is equals to the empty set may arise. In that situation, a lot of confusion appear if the uniqueness of the empty set is unknown, because showing that the empty set, which we know has no element, is equal to another different, but also with no element, makes no sense, since the empty set is just one, the $\emptyset$.

So, we can claim that there's only one set with no elements.

Proof: Suppose $A, B$ are sets with no elements. It's known that $A \subset B$, and $B \subset A$. Then, by definition of equality, $A = B$.

In this way, in order to show that some set is equals to the empty set, it's necessary only to show that the set has no elements, and, due to the uniqueness, it will be undoubtedly the empty set $\emptyset$.

Proving properties of set's operations

Defining operations on sets

In order to prove some properties of set's operations, it's worth precisely defining those operations:

• Union: $A \cup B = \{x \mid x \in A \lor x \in B\}$
• Intersection: $A \cap B = \{x \mid x \in A \land x \in B\}$
• Difference: $A - B = \{x \mid x \in A \land x \not \in B\}$
• Complement: $A^\complement = \{x \mid x \not \in A\}$

Summary

1. Comutative Laws
2. Associtative Laws
3. Distributive Laws
4. Identity Laws
5. Complement Laws
6. Double Complement Laws
7. Idempotent Laws
8. Universal Bound Laws
9. De Morgan's Laws
10. Absorption Laws
11. Complements of U and $\emptyset$
12. Set Difference Law

Proving properties

1 - Commutative Laws

a) $A \cup B = B \cup A$

Let $A, B$ be arbitrary sets.
Suppose $x \in A \cup B$.
Then, by definition, $x \in A \lor x \in B$.
Case 1: if $x \in A$, $x \in B \lor x \in A$.
Case 2: if $x \in B$, $x \in B \lor x \in A$.
So, $\forall x \in A \cup B$, $x \in B \lor x \in A$.
By definition, $x \in B \cup A$.
$\forall A,B$, if $x \in A \cup B$, then $x \in B \cup A$.

Let $A, B$ be arbitrary sets.
Suppose $x \in B \cup A$.
Then, by definition, $x \in B \lor x \in A$.
Case 1: if $x \in B$, $x \in B \lor x \in A$.
Case 2: if $x \in A$, $x \in B \lor x \in A$.
So, $\forall x \in B \cup A$, $x \in A \lor x \in B$.
By definition, $x \in A \cup B$.
$\forall A,B$, if $x \in B \cup A$, then $x \in A \cup B$.

$\forall x \in A \cup B$, $ x \in B \cup A$.
$\forall x \in B \cup A$, $x \in A \cup B$.
By definition, $A \cup B = B \cup A$.

b) $A \cap B = B \cap A$

Let $A, B$ be arbitrary sets.
Suppose $x \in A \cap B$.
Then, by definition, $x \in A \land x \in B$ $x \in B$
$x \in A$
Then, $x \in B \land x \in A$
So, $\forall x \in A \cap B, x \in B \land x \in A$
By definition, $x \in B \cap A$
$\forall A,B$, if $x \in A \cap B$, then $x \in B \cap A$.

Let $A, B$ be arbitrary sets.
Suppose $x \in B \cap A$.
Then, by definition, $x \in B \land x \in A$ $x \in A$
$x \in B$
Then, $x \in A \land x \in B$
So, $\forall x \in B \cap A, x \in A \land x \in B$
By definition, $x \in A \cap B$
$\forall A,B$, if $x \in B \cap A$, then $x \in A \cap B$.

$\forall A,B$, if $x \in A \cap B$, $x \in B \cap A$.
$\forall A,B$, if $x \in B \cap A$, $x \in A \cap B$.
By definition, $A \cap B = B \cap A$.

2 - Associative Laws

a) $(A \cup B) \cup C = A \cup (B \cup C)$

Let $A, B, C$ be arbitrary sets.
Suppose $x \in (A \cup B) \cup C$
Then $x \in (A \cup B) \lor x \in C$
Then $(x \in A \lor x \in B) \lor x \in C$.
Case 1: if $x \in A$, $x \in A \lor (x \in B \lor x \in C)$.
Then $x \in A \lor x \in A \cup B$, then $x \in A \cup (B \cup C)$.
Case 2: if $x \in B$, $x \in B \lor x \in C$. If $x \in B \lor x \in C, (x \in B \lor x \in C) \lor x \in A.$
Then $x \in B \cup C \lor x \in A$, then $x \in (B \cup C) \cup A$.
By commutative law, $x \in A \cup (B \cup C)$.
Case 3: if $x \in C$, $x \in B \lor x \in C$. If $x \in B \lor x \in C, (x \in B \lor x \in C) \lor x \in A.$
Then $x \in B \cup C \lor x \in A$, then $x \in (B \cup C) \cup A$.
By commutative law, $x \in A \cup (B \cup C)$.
$\forall A,B,C$, if $x \in (A \cup B) \cup C$, then $x \in A \cup (B \cup C)$.

Using the same reasoning, supposing $x \in A \cup (B \cup C)$ we conclude that
$\forall A,B,C$, if $x \in A \cup (B \cup C)$, then $x \in (A \cup B) \cup C$.

$\forall A,B,C$, if $x \in (A \cup B) \cup C$, then $x \in A \cup (B \cup C)$.
$\forall A,B,C$, if $x \in A \cup (B \cup C)$, then $x \in (A \cup B) \cup C$.
By definition, $(A \cup B) \cup C = A \cup (B \cup C)$.

b) $(A \cap B) \cap C = A \cap (B \cap C)$

Let $A, B, C$ be arbitrary sets.
Suppose $x \in (A \cap B) \cap C$.
Then $x \in (A \cap B) \land x \in C$.
Then $(x \in A \land x \in B) \land x \in C$.
$x \in B$.
$x \in C$.
$x \in B \land x \in C$, then $x \in B \cap C$.
$x \in A$, then $x \in A \land x \in (B \cap C)$.
By definition, $x \in A \cap (B \cap C)$.
So, $\forall A, B, C$, if $x \in (A \cap B) \cap C$, then $x \in A \cap (B \cap C)$.

Using the same reasoning, supposing $x \in A \cap (B \cap C)$ we conclude that
$\forall A, B, C$, if $x \in A \cap (B \cap C)$, then $x \in (A \cap B) \cap C$.
By definition, $(A \cap B) \cap C = A \cap (B \cap C)$.

3 - Distributive Laws

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

Let $A, B, C$ be arbitrary sets.
Suppose $x \in A \cup (B \cap C)$.
Then $x \in A \lor x \in (B \cap C).$
Case 1: if $x \in A$, $x \in A \lor x \in B$, then $x \in A \cup B$.
Also, if $x \in A$, $x \in A \lor x \in C$, then $x \in A \cup C$.
Then, if $x \in A$, $x \in A \cup B \land x \in A \cup C$.
By definition, $x \in (A \cup B) \cap (A \cup C).$
$\forall A,B,C$, if $x \in A \cup (B \cap C)$, then $x \in (A \cup B) \cap (A \cup C)$.

Let $A, B, C$ be arbitrary sets.
Suppose $x \in (A \cup B) \cap (A \cup C).$
Then $x \in (A \cup B) \land x \in (A \cup C).$
Then $(x \in A \lor x \in B) \land (x \in A \lor x \in C).$
Case 1: If $x \in A$, $x \in A \lor x \in (B \cap C)$, then $x \in A \cup (B \cap C)$.
Case 2: If $x \not \in A$, $x \in B \land x \in C$, then $x \in (B \cap C)$.
If $x \in (B \cap C)$, $x \in A \lor x \in (B \cap C)$.
By definition, $x \in A \cup (B \cap C)$.
$\forall A,B,C$, if $x \in (A \cup B) \cap (A \cup C)$, then $x \in A \cup (B \cap C)$.

$\forall A,B,C$, if $x \in A \cup (B \cap C)$, then $x \in (A \cup B) \cap (A \cup C)$.
$\forall A,B,C$, if $x \in (A \cup B) \cap (A \cup C)$, then $x \in A \cup (B \cap C)$.
By definition, $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.

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

Let $A, B, C$ be arbitrary sets.
Suppose $x \in A \cap (B \cup C)$.
Then $x \in A \land x \in (B \cup C).$
$x \in A$.
$x \in (B \cup C).$
Then, $x \in B \lor x \in C$.
Case 1: If $x \in B, x \in A \land x \in B$, then $x \in A \cap B$.
Case 2: If $x \in C, x \in A \land x \in C$, then $x \in A \cap C$.
Then, $x \in A \cap B \lor x \in A \cap C$.
By definition, $x \in (A \cap B) \cup (A \cap C).$
$\forall A,B,C$, if $x \in A \cap (B \cup C)$, then $x \in (A \cap B) \cup (A \cap C)$.

Let $A, B, C$ be arbitrary sets.
Suppose $x \in (A \cap B) \cup (A \cap C)$.
Then $x \in (A \cap B) \lor x \in (A \cap C)$.
Case 1: if $x \in (A \cap B), x \in A \land x \in B$.
If $x \in B, x \in B \lor x \in C$, then $x \in B \cup C$.
Given that $x \in A \land x \in B \cup C, x \in A \cap (B \cup C)$.
Case 2: if $x \in (A \cap C), x \in A \land x \in C$.
If $x \in C, x \in B \lor x \in C$, then $x \in B \cup C$.
Given that $x \in A \land x \in B \cup C, x \in A \cap (B \cup C)$.
$\forall A,B,C$, if $x \in (A \cap B) \cup (A \cap C)$, then $x \in A \cap (B \cup C)$.

$\forall A,B,C$, if $x \in A \cap (B \cup C)$, then $x \in (A \cap B) \cup (A \cap C)$.
$\forall A,B,C$, if $x \in (A \cap B) \cup (A \cap C)$, then $x \in A \cap (B \cup C)$.
By definition, $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

4 - Identity Laws

a) $A \cup \emptyset = A$

Let $A$ be an arbitrary set.
Suppose $x \in A \cup \emptyset$.
Then $x \in A \lor x \in \emptyset$.
Case 1: if $x \in A, x \in A$.
Case 2: if $x \not \in A, x \in \emptyset$. Absurd, then $x \in A$.
$\forall A$, if $x \in A \cup \emptyset$, then $x \in A$.

Let $A$ be an arbitrary set.
Suppose $x \in A$.
If $x \in A$, $x \in A \lor x \in \emptyset$, then $x \in A \cup \emptyset$.
$\forall A$, if $x \in A, x \in A \cup \emptyset$.

$\forall A$, if $x \in A \cup \emptyset$, then $x \in A$.
$\forall A$, if $x \in A, x \in A \cup \emptyset$.
By definition, $A \cup \emptyset = A$.

b) $A \cap U = A$

Let $A$ be an arbitrary set and $A \subset U$.
Suppose $x \in A \cap U$.
Then $x \in A \land x \in U$.
$x \in A$.
$\forall A$, if $x \in A \cap U$, then $x \in A$.

Let $A$ be an arbitraty set and $A \subset U$.
Then $\forall x \in A, x \in U$.
Suppose $x \in A$, then $x \in U$, then $x \in A \cap U$.
$\forall A$, if $x \in A$, then $x \in A \cap U$.

$\forall A$, if $x \in A \cap U$, then $x \in A$.
$\forall A$, if $x \in A$, then $x \in A \cap U$.
By definition, $A \cap U = A$.

5 - Complement Laws

a) $A \cup A^\complement = U$

Let $A$ be an arbitrary set and $A \subset U$.
Suppose $x \in A \cup A^\complement$.
Then $x \in A \lor x \in A^\complement$.
$x \in U$.
Case 1: if $x \in A$, then $x \in U$.
Case 2: if $x \in A^\complement$, $x \not \in A$, $x \in U$.
$\forall A$, if $x \in A \cup A^\complement$, then $x \in U$.

Let $A$ be an arbitrary set and $A \subset U$.
Suppose $x \in U$.
$x \in A \lor x \not \in A$.
So, $x \in A \cup A^\complement$.
$\forall A$, if $x \in U$, then $x \in A \cup A^\complement$.

$\forall A$, if $x \in A \cup A^\complement$, then $x \in U$.
$\forall A$, if $x \in U$, then $x \in A \cup A^\complement$.
By definition, $A \cup A^\complement = U$.

b) $A \cap A^\complement = \emptyset$

Let $A$ be an arbitraty set.
Suppose $x \in A \cap A^\complement$.
Then $x \in A \land x \not \in A$. Absurd.
$\forall A, \not \exists x \mid x \in A \cap A^\complement$.
By the the definition of an empty set and the uniqueness of the empty set, $A \cap A^\complement = \emptyset$.

6 - Double Complement Law

a)$(A^\complement)^\complement = A$

Let $A$ be an arbitrary set.
Suppose $x \in (A^\complement)^\complement$. Then $x \not \in A^\complement$. Then $x \in A$. $\forall A$, if $x \in (A^\complement)^\complement$, then $x \in A$.
Let $A$ be an arbitrary set.
Suppose $x \in A$.
Then $x \not \in A^\complement$.
By definition, $x \in (A^\complement)^\complement$.
$\forall A$, if $x \in A$, then $x \in (A^\complement)^\complement$.
$\forall A$, if $x \in (A^\complement)^\complement$, then $x \in A$. $\forall A$, if $x \in A$, then $x \in (A^\complement)^\complement$. By definition, $(A^\complement)^\complement = A$.

7 - Idempotent Laws

a)$A \cup A = A$

Let $A$ be an arbitrary set.
Suppose $x \in A \cup A$.
Then $x \in A \lor x \in A$.
$x \in A$.
$\forall A$, if $x \in A \cup A$, then $x \in A$.

Let $A$ be an arbitrary set.
Suppose $x \in A$.
Then $x \in A \lor x \in A$.
By definition, $x \in A \cup A$.
$\forall A$, if $x \in A$, then $x \in A \cup A$.

$\forall A$, if $x \in A \cup A$, then $x \in A$.
$\forall A$, if $x \in A$, then $x \in A \cup A$.
By definition, $A \cup A = A$.

b)$A \cap A = A$

Let $A$ be an arbitrary set.
Suppose $x \in A \cap A$.
Then $x \in A \land x \in A$.
$x \in A$.
$\forall A$, if $x \in A \cap A$, then $x \in A$.

Let $A$ be an arbitrary set.
Suppose $x \in A$.
Then $x \in A \land x \in A$.
By definition, $x \in A \cap A$.
$\forall A$, if $x \in A$, then $x \in A \cap A$.

$\forall A$, if $x \in A \cap A$, then $x \in A$.
$\forall A$, if $x \in A$, then $x \in A \cap A$.
By definition, $A \cap A = A$.

8 - Universal Bound Laws

a)$A \cup U = U$

Let $A$ be an arbitrary set and $A \subset U$.
By definition of subset, $\forall x \in A, x \in U$.
Suppose $x \in A \cup U$.
Then $x \in A \lor x \in U$.
Case 1: if $x \in A, x \in U$.
Case 2: if $x \in U, x \in U$.
$\forall x \in A \cup U, x \in U$.

Let $A$ be an arbitrary set and $A \subset U$.
By definition of subset, $\forall x \in A, x \in U$.
Suppose $x \in U$.
Then $x \in A \lor x \in U$, then $x \in A \cup U$.
$\forall x \in A, x \in A \cup U$.

$\forall x \in A \cup U, x \in U$.
$\forall x \in A, x \in A \cup U$.
By definition, $A \cup U = U$.

b)$A \cap \emptyset = \emptyset$

Let $A$ be an arbitrary set.
Suppose $x \in A \cap \emptyset$.
Then $x \in A \land x \in \emptyset$.
$x \in \emptyset$. Absurd.
Then $\not \exists x \mid x \in A \cap \emptyset$.

By the the definition of an empty set and the uniqueness of the empty set, $A \cap \emptyset = \emptyset$.

9 - De Morgan's Laws

a)$(A \cup B)^\complement = A^\complement \cap B^\complement$

b)$(A \cap B)^\complement = A^\complement \cup B^\complement$

10 - Absorption Laws

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

Let $A, B$ be arbitrary sets.
Suppose $x \in A \cup (A \cap B)$.
Then $x \in A \lor x \in (A \cap B)$.
Case 1: if $x \in A$, then $x \in A$.
Case 2: if $x \in A \cap B$, then $x \in A \land x \in B$, then $x \in A$.
$\forall A$, if $x \in A \cup (A \cap B)$, then $x \in A$.

Let $A, B$ be arbitrary sets.
Suppose $x \in A$.
Then $x \in A \lor x \in A \cup (A \cap B)$.
$\forall A$, if $x \in A$, then $x \in A \cup (A \cap B)$.

$\forall A$, if $x \in A \cup (A \cap B)$, then $x \in A$.
$\forall A$, if $x \in A$, then $x \in A \cup (A \cap B)$.
By definition, $A \cup (A \cap B) = A$.

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

Let $A, B$ be arbitrary sets.
Suppose $x \in A \cap (A \cup B)$.
Then $x \in A \land x \in (A \cup B)$.
$x \in A$.
$\forall A$, if $x \in A \cap (A \cup B)$, then $x \in A$.

Let $A, B$ be arbitrary sets.
Suppose $x \in A$.
Then $x \in A \lor x \in B$, then $x \in (A \cup B)$.
Then $x \in A \land x \in (A \cup B)$, then $x \in A \cap (A \cup B)$.
$\forall A$, if $x \in A$, then $x \in A \cap (A \cup B)$.

$\forall A$, if $x \in A \cap (A \cup B)$, then $x \in A$.
$\forall A$, if $x \in A$, then $x \in A \cap (A \cup B)$.
By definition, $A \cap (A \cup B) = A$.

11 - Complements of U and $\emptyset$

a)$U^\complement = \emptyset$

Let $A$ be an arbitrary set and $A \subset U$.
So, $\forall x \in A, x \in U$.
Suppose $x \in U^\complement$.
Then, $x \not \in U$. Absurd.
Then, $\not \exists x \mid x \in U^\complement$.

By the the definition of an empty set and the uniqueness of the empty set, $U^\complement = \emptyset$.

b)$\emptyset^\complement = U$

Suppose an arbitrary $x \in \emptyset^\complement$.
Then $x \not \in \emptyset$.
Then $x \in U$.
$\forall x \in \emptyset^\complement$, $x \in U$.

Suppose an arbitrary $x \in U$.
Then $x \not \in \emptyset$.
Then $x \in \emptyset^\complement$.
$\forall x \in U$, then $x \in \emptyset^\complement$.

$\forall x \in \emptyset^\complement$, $x \in U$.
$\forall x \in U$, then $x \in \emptyset^\complement$.
By definition, $\emptyset^\complement = U$.

12 - Set Difference Law

a)$A - B = A \cap B^\complement$

Let $A, B$ be arbitrary sets.
Suppose $x \in A - B$.
Then $x \in A \land x \not \in B$.
$x \in A$.
By definition, if $x \not \in B$, then $x \in B^\complement$.
Then, $x \in A \land x \in B^\complement$, then $x \in A \cap B^\complement$.
$\forall A, B$, if $x \in A - B$, then $x \in A \cap B^\complement$.

Let $A, B$ be arbitrary sets.
Suppose $x \in A \cap B^\complement$.
Then, $x \in A \land x \in B^\complement$.
If $x \in B^\complement$, then $x \not \in B$.
$x \in A$.
$If x \in A \land x \not \in B$, by definition, $x \in A - B$.
$\forall A, B$, if $x \in A \cap B^\complement$, $x \in A - B$.

$\forall A, B$, if $x \in A - B$, then $x \in A \cap B^\complement$.
$\forall A, B$, if $x \in A \cap B^\complement$, $x \in A - B$.
By definition, $A - B = A \cap B^\complement$.