The proof of (P ←→ q) = (P ∨ q) ∧ (P ∨ q) in Discrete Mathematics Yes [proved by basic equivalent formula]

┐(P←→Q)
＝┐((P→Q)∧(Q→P))
＝┐((┐P∨Q)∧(┐Q∨P))
＝┐(┐P∨Q)∨┐(┐Q∨P)
＝(P∧┐Q)∨(Q∧┐P)
＝(P∨Q)∧(P∨┐P)∧(┐Q∨Q)∧(┐Q∨┐P)
＝(P∨Q)∧1∧1∧(┐Q∨┐P)
＝(P∨Q)∧(┐P∨┐Q)

What is the self - complementing graph in discrete mathematics? How many are the non - term self - complementing graphs with 5 vertices

Complement graph: given a graph G, all the nodes in G and all the added edges that make g a complete graph are complement graphs
Self complementary graph: a graph is self complementary if it is isomorphic to its complement
There should be two self complementary graphs with five vertices. Explain and draw a graph with reference to the definition

Given the degree of each vertex in a simple graph, judge whether the graph can be drawn
For example, if the value in the alternative answer represents the degree of each vertex in a simple graph, what can draw the graph is ()
(A)(1,2,2,3,4,5) (B)(1,2,3,4,5,5)\x05 (C)(1,1,1,2,3) (D)(2,3,3,4,5,6).

First, the sum of degree is even, then judge whether it is a simple graph. Method: delete the point with the largest degree in turn, recursively, and finally determine whether it is a simple graph

1. It is proved that in a simple undirected graph G with n vertices, at least two vertices have the same degree

If n vertex degrees are d (XI) (1 ≤ I ≤ n), then d (XI) can take 0,1,2... And N-1 can take n different values. If D (XI) = 0, then d (XI) = NN and D (XI) can take n-1 different values. According to pigeonhole principle, there must be d (XM) = D (xn), that is, there must be vertex points with the same degree. If D (XI) = n, then d (x) can not exist

How to draw the following four diagrams of discrete mathematical problems? Can you explain them in detail? Thank you

For example, R (R) is a reflexive closure and must be supplemented with ARA, so there are four self rings

On the problem of set in Discrete Mathematics
Is a finite set a countable set?
Let a be a finite set and B be a countable set. Why is the Cartesian product of a and B an infinite set?

Let n be all positive integers and N = {1,2,3 If there is a positive integer n, then n is called a finite set. But can you count how many elements there are in the set? Of course, No. an empty set is also considered a finite set. But an empty set has elements. Let a be a finite set and B be a finite set

On the number of set transitive relations in Discrete Mathematics
If a set has n elements, how many transitive relationships are there on the set?

Is reflexivity directed or undirected

&Let d be the set of lines on the same plane, and / / denote the parallel relationship between two lines, and ⊥ denote the vertical relationship between two lines, then / / ^ 2 & nbsp; = & nbsp; / / & nbsp; & nbsp;, ⊥ ^ 2 = & nbsp; & nbsp; / / & nbsp; & nbsp;, ⊥ ^ 3 = & nbsp; & nbsp; & nbsp; ⊥

In the same plane, a, B, C and D respectively represent the lines that are not coincident with each other in this plane
1. If a ∥ B, C ∥ B, then a ∥ B ∥ C, so a ∥ C, that is / / ^ 2 =//
2. If a ⊥ B, C ⊥ B, then a ⊥ C, that is, ⊥ 2 =//
3. If a ∥ C, C ⊥ D, then a ⊥ D, and ∵ from 2, we can get ⊥ 2 = / / that is, / = ⊥ 2
∴ ⊥^3＝⊥
The precondition is very important, one is that it is in the same plane, and the other is that the four lines do not coincide with each other!

Let a = {a, B, C}, B = {1,2}, then:
Find B ^ a
Find a ^ a

There are 2 ^ 3 = 8 functions from a to B
f1:a→1,b→1,c→1
f2:a→1,b→1,c→2
f3:a→1,b→2,c→1
f4:a→1,b→2,c→2
f5:a→2,b→1,c→1
f6:a→2,b→1,c→2
f7:a→2,b→2,c→1
f8:a→2,b→2,c→2
B^A={f1,f2,f3,f4,f5,f6,f7,f8},|B^A|=8
There are 3 ^ 3 = 27 functions from a to a
g1:a→a,b→a,c→a
g2:a→a,b→a,c→b
.
Write the rest yourself
A^A={g1,g2,...,g27},|A^A|=27

I'd like to ask the judgment of reflexivity and anti reflexivity, symmetry and antisymmetry in discrete mathematics
(1) If any x (x ∈ a → & lt; X, X & gt; ∈ R), then R is said to be reflexive on a
(2) If any x (x ∈ a → & lt; X, X & gt; R), then R is said to be anti reflexive on a
(1) If any x, any y (x, y ∈ a ∧ & lt; X, Y & gt; ∈ R → & lt; y, X & gt; ∈ R), then R is called symmetric relation on A. (2) if any x, any y (x, y ∈ a ∧ & lt; X, Y & gt; ∈ R ∧ & lt; y, X & gt; ∈ R → x = y), then R is called antisymmetric relation on a
Above is the definition
Let a = {1,2,3}, R1 = {& lt; 1,1 & gt;, & lt; 2,2 & gt;} I know that he is neither reflexive nor anti reflexive,
That means that if it is reflexive, it must contain IA (& lt; 1,1 & gt;, & lt; 2,2 & gt;, & lt; 3,3 & gt;),
Does any x described by definition 1 contain all the numbers in a?
---------------------------------------------
R1＝{<1,1>,<2,2>}R2＝{<1,1>,<1,2>,<2,1>}
R3＝{<1,2>,<1,3>}R4＝{<1,2>,<2,1>,<1,3>}
R1 is both symmetric and antisymmetric. R2 is symmetric but not antisymmetric
R3 is antisymmetric but not symmetric. R4 is neither symmetric nor antisymmetric
It's written in the book, and then
2. Why don't R1 and R2 contain element 3? Isn't it true that any X and Y belong to a? If I misunderstand it, why should question 1 contain all the elements of a?

In the definitions of these relational properties in the book, the values of the variables X and y of the first-order logic formula are total individual fields, so there are restrictions on X ∈ A and Y ∈ a in the scope. In fact, we only consider them in set a, so these definitions can completely remove the restrictions on X ∈ A and Y ∈ a

The proof of (P ←→ q) = (P ∨ q) ∧ (P ∨ q) in Discrete Mathematics Yes [proved by basic equivalent formula]