The logic

by
Nittaya Noinan
Kanchanapisekwittayalai Phetchabun School

Proposition or Statement
Statement is sentence that is either
true or false only .

Example
The following sentences are statement.
1. Cheangmai is not in the south of Thailand. (True)
2. 9 3 (True)
3. 17 + 8 25 (False)
4. is Rational Numbers (False)
5. Empty Set is subset of all of set (True)

Example
The following sentences are not statement.
1. Is it rained ?
2. Don’t move your body.
3. Help me !
4. Please open the door.
5. Waw ! So beautiful.
6. Get out !
7. I wish to travel so much.

Connectives
1. And
2. Or
3. If…that… or Implies
4. If and only if or equivalent to
5. Negation or Xor

Connectives
1. And the symbol is “ ”
F
F
F
T
P Q
FF
TF
FT
TT
QP

Connectives
1. And symbol is “ ”
Example
1) is irrational number and is odd number
solve
Give p is “ is irrational number” has truth –value is true
q is “ is odd number” has truth –value is true
Then we can write be symbol is p q
Thus this statement has truth – value is true
Answer True

Connectives
1. And symbol is “ ”
Example
2) 0 is integer number and 2.4 is even number
solve
Give p is “ 0 is integer number” has truth –value is true
q is “ 2.4 is even number” has truth –value is false
Thus this statement has truth – value is false
Answer False

Connectives
2. Or the symbol is “ ”
F
T
T
T
P Q
FF
TF
FT
TT
QP

Connectives
2. Or symbol is “ ”
Example
1) 2 is even number or 23 is odd number
solve
Give p is “ 2 is even number” has truth –value is true
q is “ 23 is odd number” has truth –value is false
Answer True

Connectives
2. Or symbol is “ ”
Example
2) is rational number or is interger number
solve
Give p is “ is rational number” has truth –value is false
q is “ is integer number” has truth –value is false
Answer False

Connectives
3. If…then… the symbol is “ ”
T
T
F
T
P Q
FF
TF
FT
TT
QP

Connectives
3. If …then… symbol is “ ”
Example
1) If 2 + 3 = 3 + 2 then 6(2 + 3) = 6(3 + 2)
solve
Give p is “2 + 3 = 3 + 2 ” has truth –value is true
q is “6(2 + 3) = 6(3 + 2)” has truth –value is true
Answer True

Connectives
3. If …then… symbol is “ ”
Example
2) If 5 is odd number then 54 is even number
solve
Give p is “ 5 is odd number” has truth –value is true
q is “ 54 is even number” has truth –value is false
Answer False

Connectives
4. If and only if the symbol is “ ”
T
F
F
T
P Q
FF
TF
FT
TT
QP

Connectives
4. If and only if symbol is “ ”
Example
1) 7 is divided by 2 if and only if 7 is even number
solve
Give p is “ 7 is divided by 2 ” has truth –value is false
q is “7 is even number” has truth –value is false
Answer True

Connectives
4. If and only if symbol is “ ”
Example
2) 2 < 3 if and only if
solve
Give p is “ 2 < 3 ” has truth –value is true
q is “ ” has truth –value is false
Answer False

Connectives
5. Negation the symbol is “ ”
TF
FT
PP

Connectives
5. Negation symbol is “ ”
Example
1) 2 + 3 = 5
solve
Give p is “ 2 + 3 = 5 ” has truth –value is true
Negation of this statement is “2 + 3 ≠ 5”
has truth –value is false
Then we can write be symbol is p
Thus Negation of this statement has truth – value is false
Answer False

Connectives
5. Negation symbol is “ ”
Example
2) Spider is not insect
solve
Give p is “Spider is not insect” has truth –value is false
Negation of this statement is “Spider is insect”
has truth –value is true
Then we can write be symbol is p
Thus Negation of this statement has truth – value is true
Answer True

Truth – value of Statement
Example Find the Truth – value of the fallowing statements
1) Cheangmai and Thonburi used to capital of Thailand
Solve
Give p is Cheangmai used to capital of Thailand
Give q is Thonburi used to capital of Thailand
Since p has the truth – value is false
q has the truth – value is true
Answer False

Example
2) Give a , b and c be statements have truth – value is true ,
true and false following order Find the Truth – value of the
fallowing statements ( a b ) c
Solve
a b has the truth – value is true and c is false
Thus ( a b ) c has truth – value is true
Answer True

Example
3) Find the Truth – value of the fallowing statements ( a b )
When a , b be statements all have truth – value is true
Solve
( a b )
T T
F
F
T
Answer True

Example
4) Give p , q , r and s be statements have truth – value is true
, false , false and true following order Find the Truth – value
of the fallowing statements [( p q ) r] (p s)
Solve
[( p q ) r] (p s)
T F F T T
F T T
T
T
Answer True

Order to find the Truth – value of Statement
1. Braces
2.
3. “ ” or “ ”
4.
5.

Truth Table
You can see p , q and r be atomic statement from
the statement that has connective such as p , p q
, p q , (p q) r , (p q) ect.
When you don’t give the truth – value p , q and r be
variable instead of statement. And we call p , p q
, p q , (p q) r , (p q) are statement pattern.
Since p , q and r be variable instead of statement then
the truth – value of statement pattern have to give the
atomic statement all cases.

Truth Table
* If there is one statement (p) then the cases of the
truth – value be 2 cases
* If there is two statements (p,q) then the cases of the
* If there is three statements (p, q ,r) then the cases of the
* If there is four statements (p,q,r,s) then the cases of the
Thus if there is n statements then the cases of the
truth – value be 2n cases

Truth Table
Example.
1) Give p and q be statements . Make the truth table of
(p q) ( p q)
Solve
p q P q p q p q (p q) ( p q)
T T T F F F F
T F F F T F T
F T T T F F F
F F T T T T T

Truth Table
Example.
[(p q) (p q)]
Solve
p q P q q p q (p q) (p q)] [(p q) (p q)]
T T T F T T F
T F F T T T F
F T F F F T F
F F T T T T F

Truth Table
Example.
3) Give p , q and r be statements . Make the truth table of
(p q) ( r q)
Solve
p q r p q r r q (p q) ( r q)
T T T T F T T
T T F T T T T
T F T F F F T
T F F F T T T
F T T T F T T
F T F T T T T
F F T T F F F
F F F T T T T

Truth Table
Example.
4) Give p ,q and r be statements . Make the truth table of
p ( q) and (p q)
Solve
p q q p ( q) (p q) (p q)
T T F F T F
T F T T F T
F T F F T F
F F T F T F
The following the truth table you can observe p ( q) and (p q)
is the same case by case

Equivalent pattern
Observe the following examples.
p q and q p
Solve
p q p q (p q) q p
T T F F T T
T F F T F F
F T T F T T
F F T T T T
The following the truth table you can observe p q and q p
is the same case by case .Then we can say p q equivalent q p
Thus we can write be symbol is p q q p

Equivalent pattern
(p q) and p q
Solve
p q p q p q (p q) P q
T T F F T F F
T F F T T F F
F T T F T F F
F F T T F T T
The following the truth table you can observe (p q) and p q
is the same case by case .Then we can say (p q) equivalent p q
Thus we can write be symbol is (p q) p q

Equivalent pattern
p q and p q
Solve
p q p p q P q
T T F F T
T F F F F
F T T T T
F F T F T
The following the truth table you can observe p q and p q
is not the same case by case .
Then we can say p q is not equivalent p q

Equivalent pattern
4) Give p and q be statements . Consider (p q) (q p)
equivalent p q
Solve
p q p q q p (p q) (q p) P q
T T T T T T
T F F T F F
F T T F F F
F F T T T T
The following the truth table you can observe (p q) (q p) and p q
is the same case by case .Then we can say (p q) (q p) equivalent p q
Thus we can write be symbol is (p q) (q p) p q

Equivalent pattern
5) Consider the two contents that be equivalent or not.
“if 82 is even number then 8 is even number”
“ if 8 is not even numbr then 82 is not even number”
Solve
Give p be 82 is even number
q be 8 is even number
Then p q instead of “if 82 is even number then 8 is even number”
q p instead of “ if 8 is not even numbr then 82 is not even number”
Thus from the example 1) we got p q equivalent q p
Answer The two contents that be equivalent

Equivalent pattern
Symbol of equivalent pattern is “ ”
The following two statements are equivalent.
1) ( p) p
2) p q q p
3) p q q p
4) p q q p
5) p (q r) (p q) r
6) p (q r) (p q) r
7) p (q r) (p q) (p r)
8) p (q r) (p q) (p r)
9) (p q) p q
10) (p q) p q

Equivalent pattern
11) (p q) p q
12) p q p q
13) p q q p
14) p q p q
15) p q (p q) (q p)
16) p p p
17) p p p
18) (p q) r (p r) (q r)
19) (p q) r (p r) (q r)
20) p (q r) (p q) (p r)
21) p (q r) (p q) (p r)

Equivalent pattern
22) p T p
T q q
23) p F p
F q q
24) T q q
p F p
25) T q q
p T p
F q q
p F p

Tautology
Definiton.
The truth – value of statement pattern is true all cases
is called Tautology.

Tautology
Example.
1) Consider the truth – value of statement pattern
[(p q) p] q
p q p q (p q) p [(p q) p] q
T T T T T
T F F F T
F T T F T
F F T F T
You can observe [(p q) p] q has the truth – value is
true all cases. Then we can say [(p q) p] q is Tautology.

Tautology
Example.
(p q) ( q p) be Tautology or not.
p q p q p q q p (p q) ( q p)
T T F F T T T
T F F T F F T
F T T F T T T
F F T T T T T
You can observe (p q) ( q p) has the truth – value is
true all cases. Then we can say (p q) ( q p) is Tautology.

Tautology
Example.
[(p q) p] q be Tautology or not.
p q p q p q (p q) p [(p q) p] q
T T F F T F T
T F F T F T T
F T T F T T F
F F T T T T T
You can observe [(p q) p] q has the truth – value is true
not all cases. Then we can say (p q) ( q p) is not
Tautology.

Tautology
By the ways we can check which the statement pattern
is tautology when the statement pattern has connective is
“if… then…” by method assume contradiction.
That mean assume the statement pattern be false. It has
the truth – value is false only one case. ( T F F )
If it is contradition from the assume before. Then the
statement pattern be Tautology.

Tautology
Example.
1) Give p and q be statements . Consider the statement
pattern [(p q) p] q is Tautology or not.
Solve.
assume that [(p q) p] q is false.
[(p q) p] q
F
T F
T T T
T T
Now you can observe we assume [(p q) p] q is false and
then it is not contradiction .
Thus the truth – value of p is T and the truth – value of q is T
make the truth – value [(p q) p] q is false .
Answer [(p q) p] q is not Tautology.

Tautology
Example.
2) Give p and q be statements . Consider the statement pattern
(p q) (q p) is Tautology or not.
Solve.
assume that (p q) (q p) is false.
(p q) (q p)
F
T F
T T F F
Now you can observe we assume (p q) (q p) is false and
then it is contradiction .
Thus the truth – value of (p q) (q p) can’t be false .
Answer (p q) (q p) is Tautology.
Contradiction

Tautology
Example.
1) Give p and q be statements . Consider the statement
pattern ( q p) (q p) is Tautology or not.
Solve.
assume that ( q p) (q p) is false.
( q p) (q p)
F
T F
F T T F
T T
Now you can observe we assume ( q p) (q p) is false and
then it is not contradiction .
Thus the truth – value of p is T and the truth – value of q is T
make the truth – value ( q p) (q p) is false .
Answer ( q p) (q p) is not Tautology.

Tautology
Example.
4) Give p ,q and r be statements . Consider the statement
pattern [(p q) (q r)] (p r) is Tautology or not.
Solve.
assume that [(p q) (q r)] (p r) is false.
[(p q) (q r)] (p r)
F
T F
T T T F
T T T T
Now you can observe we assume [(p q) (q r)] (p r)
is false and then it is contradiction .
Thus the truth – value of [(p q) (q r)] (p r) can’t be
false .
Answer [(p q) (q r)] (p r) is Tautology.
Contradiction

Argument
Argument that there are statements be causes
(P1, P2, P3,…,Pn) and can be conclusion (C) .
Thus argument has 2 parts such that causes and
coclusion.
We can do argument by use Connective “ ” between
statement that are causes. And use connective “ ”
between causes and conclusion.
(P1 P2 P3 … Pn) C
There are 2 types of argument that are valid and
invalid.

Argument
(P1 P2 P3 … Pn) C
There are 2 types of argument that are valid and
invalid.
* (P1 P2 P3 … Pn) C be tautology that mean
this argument is valid
* (P1 P2 P3 … Pn) C be not tautology that mean
this argument is invalid

Argument
Example.
1) Give p and q be statement. Consider the following
argument is valid or invalid.
cause 1. p q
2. p
conclusion q
Solve.
[(p q) p ] q
F
T F
T T
T T
Since [(p q) p ] q is Tautology Thus this argument is valid
Contradiction

Argument
Example.
2) Give p , q and r be statement. Consider the following
argument is valid or invalid.
cause 1. p q
2. q r
3. r
conclusion p
Solve.
[(p q) ( q r ) r ] p
F
T T F
T T T T
T T F F F
T
Since [(p q) p ] q is not Tautology Thus this argument is invalid

Argument
Example.
3) Consider the following argument is valid or invalid.
cause 1. If it rained then the roof is wet
2. the roof is not wet
conclusion it is not rained
Solve.
Give p be it rained
q be the roof is wet
Then write the symbol.
cause 1. p q
2. q
conclusion p
Thus we can write [(p q) q ] p

Argument
[(p q) q ] p
F
T F
T T T
T T F
Since [(p q) p ] q is Tautology
Thus this argument is valid
Contradiction

Argument
Example.
4) Consider the following argument is valid or invalid.
cause 1. If Dang get bonus then Dang deposit to the bank
2. if Dang deposit to the bank then Dang doesn’t have debt
conclusion If Dang doesn’t get bonus then Dang doesn’t have debt
Solve.
Give p be Dang get bonus
q be Dang deposit to the bank
r be Dang doesn’t have debt
Then write the symbol.
cause 1. p q
2. q r
conclusion p r
Thus we can write [(p q) (q r )] ( p r)

Argument
[(p q) (q r )] ( p r)
F
T F
T T T F
F F F F F
Since [(p q) p ] q is not Tautology
Thus this argument is invalid

Open sentence
Open sentence is sentence that
there is variable when you instead
variable by the member in the universe
then you get statement.

Example
The following sentences are Open sentence.
1. She is student in Kanchanapisekwittayalai school.
2. x 3
3. x2 - 2x + 8 = (x – 4)(x + 2)
4. x.x = x + x
5. x { 1, 2, 3 }
6. If x be natural number then is not real number.

Example
The following sentences are not Open sentence.
1. Is she your sister?
2. Don’t move your body.
3. X2 + 1
4. 2 { 1, 2, 3 }
5. 2x – 1

Quantifier
There are 2 Quantifier in Mathematics.
Such that
1. “For all” the symbol is “ ”
2. “For some” the symbol is “ ”

Quantifier
Symbol that we have to know.
1) x be For all of x
2) x be For some of x
3) U be Universe
4) R be Set of real number
5) Q be Set of rational number
6) I or Z be Set of integer number
7) N be Set of natural number

Quantifier
Example. Write the sentence be symbol.
1) For all of x that x + x = 2x
Answer. x[x + x = 2x] ,U = R
2) there are x that x + 0 = 2x
Answer. x[x + 0 = 2x] ,U = R
3) For all of x if x is integer number then x is real number
Answer. x[x I x R] ,U = R or you can say
all interger number be real number.

Quantifier
Example. Write the sentence be symbol.
4) For all real number be integer number.
Answer. x[x I ] ,U = R
5) For some integer number square equal 1
Answer. x[x I x2 = 1] ,U = R
if the problem doesn’t refer to universe we
should to understand that mean U = R

Quantifier
Example. Write the symbol besentence.
1) x [x I x < 5] ,U = R
Answer. There are x is integer number and less than 5
2) x[x I x < 5]
Answer. For all of x if x is integer number then less than 5
3) y x [x2 + y2 = 8]
Answer. For all of y and there are x that x2 + y2 = 8

Quantifier
Example. Write the symbol besentence.
4) y x [x2 + y2 = 8]
Answer. For some of y and all of x that x2 + y2 = 8
5) x y[x + y R]
Answer. For all of x and y that sum of x + y be real
number

The truth – value of Quantifier sentence.
Consider the truth – value of Quantifier sentence
that there 3 parts .
1. Quantifier
2. Open sentence
3. Universe
The symbol of the truth – value of Quantifier sentence
by instead Open sentence that has x be variable be P(x)
Thus the truth – value of Quantifier sentence can write be
symbol such that
1. x[P(x)] when universe is U
2. x[P(x)] when universe is U

Definition.
x[P(x)] has the truth – value is true iff when
instead all of x in universe to P(x) then all of
cases be true.
x[P(x)] has the truth – value is false
iff when instead some of x in universe
to P(x) then there are some of cases
be false.

Definition.
x[P(x)] has the truth – value is true iff when
instead some of x in universe to P(x) then
the cases be true.
x[P(x)] has the truth – value is false iff
when instead all of x in universe to P(x)
then there are all of cases be false.

Example. Find the truth – value of Quantifier sentences.
1) x[x < 5] , U = {0, 1, 2, 3 }
solve. Give P(x) = x < 5
P(0) = 0< 5 (True)
P(1) = 1< 5 (True)
P(2) = 2< 5 (True)
P(3) = 3< 5 (True)
P(4) = 4< 5 (True)
Then instead all of x in U to P(x) is true all cases.
Thus x[x < 5] , U = {0, 1, 2, 3 } has truth – value is True.

Example. Find the truth – value of Quantifier
sentences.
2) x[x < 5] , U = I
then P(8) = 8 < 5 (False)
Then instead some of x in U to P(x) is false
some case.
Thus x[x < 5] , U = I has truth – value is False.

Example. Find the truth – value of Quantifier
sentences.
3) x[x < 5] , U = I
then P(6) = 6 < 5 (True)
Then instead some of x in U to P(x) is true
some case.
Thus x[x < 5] , U = I has truth – value is True.

4) x[x < 5] , U = { 6, 7, 8 }
P(6) = 0< 5 (False)
P(7) = 1< 5 (False)
P(8) = 2< 5 (False)
Then instead all of x in U to P(x) is false all cases.
Thus x[x < 5] , U = { 6, 7, 8 } has truth – value is False.

5) x[(x < 0) (x2 > 0)] , U = {-1, 0, 1 }
solve. Give P(x) = (x < 0) (x2 > 0)
P(-1) = -1 < 0 (-1)2 > 0 (True)
P(0) = 0 < 0 (-1)2 > 0 (True)
P(1) = 1 < 0 (-1)2 > 0 (True)
Then instead all of x in U to P(x) is true all cases.
Thus x[(x < 0) (x2 > 0)] , U = {-1, 0, 1 }
has truth – value is True.

6) x[(x < 0) x(x2 > 0)] , U = {-1, 0, 1 }
solve. Give P(x) = (x < 0)
P(1) = 1 < 0 (False)
Thus x[(x < 0) , U = {-1, 0, 1 } is False
Give P(x) = (x2 > 0)
P(0) = 02 > 0 (False)
Thus x[(x2 > 0)], U = {-1, 0, 1 } is False
Hence x[(x < 0) x(x2 > 0)] be F F that is True.

7) x[(x < 0) (x - 1 = 0)] , U = {-1, 0, 1 }
solve. Give P(x) = (x < 0) (x - 1 = 0)
P(-1) = (-1 < 0) (-1 - 1 = 0) (False)
P(0) = (0 < 0) (0 - 1 = 0) (False)
P(1) = (1 < 0) (1 - 1 = 0) (False)
Then instead all of x in U to P(x) is false all cases.
Thus x[(x < 0) (x - 1 = 0)] , U = {-1, 0, 1 }
has truth – value is False.

8) x[(x < 0) x (x - 1 = 0)] , U = {-1, 0, 1 }
solve. Give P(x) = (x < 0)
P(-1) = -1 < 0 (True)
Thus x [(x < 0) , U = {-1, 0, 1 } is True
Give P(x) = (x – 1 = 0)
P(1) = (1 – 1 = 0) (True)
Thus x[(x – 1 = 0)], U = {-1, 0, 1 } is True.
Hence x[(x < 0) x (x - 1 = 0)] be T T that is True.

Equivalent and Negation of
Quantifier sentence.
Let p and q be statement
P(x) and Q(x) be Open sentence
x[P(x)] , x[P(x)] , x[Q(x)] and x[Q(x)]
be Quantifier sentence.
Then equivalent is the same together.

Equivalent and Negation of
Quantifier sentence.
Example.
statement Open sentence
p q q p P(x) Q(x) Q(x) P(x)
p q q p P(x) Q(x) Q(x) P(x)
(p q) p q (P(x) Q(x)) P(x) Q(x)
(p q) p q (P(x) Q(x)) P(x) Q(x)

Equivalent of Quantifier sentence.
Example.
1. x[P(x) Q(x)] x[ P(x) Q(x)]
2. x[P(x) Q(x)] x[ Q(x) P(x)]
3. x[ P(x) Q(x)] x[P(x) Q(x)]
4. x[ (P(x) Q(x))] x[ P(x) Q(x)]

Negation of Quantifier sentence.
x[P(x)] x[ P(x)]
x [P(x)] x[ P(x)]
Example. Find the Negation of quantifier sentence.
1. x[x + 3 > 5]
Answer x[ x + 3 5]

2. x[x - 0 = 4]
Answer x[ x – 0 ≠ 4]
3. x[x > 0] x[x2 < 0]
Solve. [ x[x > 0] x[x2 < 0]]
x[x > 0] x[x2 < 0]
x[x 0] x[x2 0]
Answer x[x 0] x[x2 0]

4. x[x ≠ 0] x[x ≠ 0]
Solve. [ x[x ≠ 0] x[x ≠ 0]]
x[x ≠ 0] x[x ≠ 0]
x[x ≠ 0] x[x = 0]
Answer x[x ≠ 0] x[x = 0]

5. x[P(x) Q(x)]
Solve. [ x[P(x) Q(x)]]
x[ [P(x) Q(x)]]
x[P(x) Q(x)
Answer x[P(x) Q(x)

6. There are real number x be rational
number but not be integer number.
Solve. Give P(x) be x is rational number
Q(x) be x is integer number
then x[P(x) Q(x)]
thus x[P(x) Q(x)]
x[ [P(x) Q(x)]]
x[ [P(x) Q(x)]]
Answer For all real number x is not necessary be
rational number all or have to be integer number.

The truth – value of two Quantifier sentences.
We can write the truth – value of two quantifier sentences
to be 8 patterns such that
x y[P(x,y)] x y[Px,y)] x y[P(x,y)] x y[P(x,y)]
y x[P(x,y)] y x[Px,y)] y x[P(x,y)] y x[P(x,y)]

Definition.
x y[P(x,y)] has the truth – value is true iff instead
x and y with a and b all in universe
then P(a,b) is true all.
x y[P(x,y)] has the truth – value is false iff instead
x and y with a and b some in universe
then P(a,b) is false.

x y[P(x,y)] or y x [P(x,y)]
Example.
1) Give U = { -1 , 0 , 1 } Find the truth – value following
Problems.
1.1 x y[xy < 2]
Solve.
Give P(x,y) be xy < 2
Then we get P(-1,-1) be (-1)(-1) < 2 (True)
P(-1,0) be (-1)(0) < 2 (True)
P(-1,1) be (-1)(1) < 2 (True)
P(0,-1) be (0)(-1) < 2 (True)
P(0,0) be (0)(0) < 2 (True)
P(0,1) be (0)(1) < 2 (True)
P(1,-1) be (1)(-1) < 2 (True)
P(1,0) be (1)(0) < 2 (True)
P(1,1) be (1)(1) < 2 (True)
Thus x y[xy < 2] is true.

Example.
Problems.
1.2 x y[x + y < 2]
Solve.
Give P(x,y) be x + y < 2
Then you can see P(1, 1) be 1 + 1 < 2 (False)
Thus x y[xy < 2] is false.

Example.
2) Find the truth – value of the following sentence.
“For all real number x and y that x + y = y + x”
Solve.
We can write to be symbol that is x y[x + y = y + x]
We can use the lesson before about property of real number
that is commutative addition of real number then x + y = y + x
always.
Hence x y[x + y = y + x] is true

Definition.
x y[Px,y)] has the truth – value is true iff instead
x and y with a and b some in universe
then P(a,b) is true.
x y[Px,y)] has the truth – value is false iff instead
x and y with a and b all in universe
then P(a,b) is false all.

x y[Px,y)] or y x[Px,y)]
Example.
Problems.
1.1 x y[2x + y = 2]
Solve.
Give P(x,y) be 2x + y = 2
Then we get P(1,0) be 2(1)+0 = 2 (True)
Thus x y[Px,y)] is true.

x y[Px,y)] or y x[Px,y)]
Example.
Problems.
1.2 x y[x + y > 2]
Solve.
Give P(x,y) be x + y > 2
Then we get P(-1,-1) be (-1) + (-1) > 2 (False)
P(-1,0) be (-1) + (0) > 2 (False)
P(-1,1) be (-1) + (1) > 2 (False)
P(0,-1) be (0) + (-1) > 2 (False)
P(0,0) be (0) + (0) > 2 (False)
P(0,1) be (0) + (1) > 2 (False)
P(1,-1) be (1) + (-1) > 2 (False)
P(1,0) be (1) + (0) > 2 (False)
P(1,1) be (1) + (1) > 2 (False)
Thus x y[x + y > 2] is False.

Definition.
x with all in universe then y[P(x,y)] is true.
x with a some in universe then y[P(x,y)]
is false.

x y[P(x,y)] or y x[P(x,y)]
Example.
Problems.
1.1 x y[x + y = 0]
Solve.
Give P(x,y) be x + y = 0
Then we instead x = -1 we can see y[x + y = 0] that is y = 1
Hence -1 + 1 = 0 .Thus is true.
Then we instead x = 0 we can see y[x + y = 0] that is y = 0
Hence 0 + 0 = 0 .Thus is true.
Then we instead x = 1 we can see y[x + y = 0] that is y = -1
Hence 1 + (-1) = 0 .Thus is true.
Thus x y[x + y = 0] is true.

Example.
Problems.
1.2 x y[x < y]
Solve.
Give P(x,y) be x < y
Then we instead x = 1 we can see y[ 1 < y]
Then give y = -1 .Thus 1 < -1 is false
y = 0 . Thus 1 < 0 is false
y = 1 . Thus 1 < 1 is false
Thus x y[x < y] is false.

Example.
2) Give U is set of real number .Find the truth – value
following Problems.
2.1 x y[x + y = 1]
Solve.
Then we instead x = a for a be all real number.
Then we get y[x + y = 1] .
Thus we can find y = a – 1 all in real number.
That a + (a – 1) = 0
Thus x y[x + y = 1] is true.

Example.
2) Give U is set of real number .Find the truth – value
following Problems.
2.2 x y[x = y2]
Solve.
Give P(x,y) be x = y2
Then we instead x = -1 thus we get y[ -1 = y2]
Hence we can’t find y2 = -1 all in real number.
Thus x y[x = y2] is false.

Definition.
x with a some in universe then y[P(x,y)]
is true.
x with a all in universe then y[P(x,y)]
is false.

Example.
Problems.
1.1 x y [x y]
Solve.
Give P(x,y) be x y
Then we instead x = -1 we can see y[ -1 y]
Hence y = -1 we get -1 -1 (True)
y = 0 we get -1 0 (True)
y = 1 we get -1 1 (True)
Thus x y [x y] is true.

Example.
Problems.
1.2 x y [x + y = 0]
Solve.
Then we instead x = -1 we can see y[ -1 + y = 0] is false.
Then we instead x = 0 we can see y[ 0 + y = 0] is false.
Then we instead x = 1 we can see y[ 1 + y = 0] is false.
Thus x y [x + y = 0] is false

Example.
2) Give U is set of real number. Find the truth – value
following Problems.
2.1 x y [x + y = y]
Solve.
Give P(x,y) be x + y = y
Then we instead x = 0 we can see y[ 0 + y = y]
Hence y = b that is all real number is true.
Thus x y [x + y = y] is true.

Example.
following Problems.
2.2 x y [x y]
Solve.
Give P(x,y) be x y
Then we instead x = a when a is real number.
we can see y [a y] is false because y can be a + 1.
Thus x y [x y] is false.

Repeat Lesson.
Example.
1) Give U is set of natural number. Find the truth – value
following Problems.
1.1 x y [y = x + 1]
Solve.
Give P(x,y) be y = x + 1
Then we instead x = a when a is natural number.
we can see y[y = a + 1] is true because y can be a + 1.
Thus x y [y = x + 1] is true.

Repeat Lesson.
Example.
1) Give U is set of natural number. Find the truth – value
following Problems.
1.2 y x [y = x + 1]
Solve.
Give P(x,y) be y = x + 1
Then we instead y = 1.
we can’t see x [ 1 = x + 1] because the natural number
is least be 1
Thus y x [y = x + 1] is false.

Repeat Lesson.
Example.
following Problems.
2.1 x y [x + y = 0]
Solve.
Then we instead x = a when a is real number.
we can see y[ a + y = 0] is true because y can be –a .
Thus x y [x + y = 0] is true.

Repeat Lesson.
Example.
following Problems.
2.2 y x [x + y = 0]
Solve.
Then we instead y = b when a is real number.
we can see x [x + b = 0] is false because x = -b only
not can be another.
Thus y x [x + y = 0] is false.

The logic

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