The logic

The Logicby

Nittaya Noinan Kanchanapisekwittayalai Phetchabun

School

Mathematics

Proposition or Statement

Statement is sentence that is eithertrue 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. And2. Or3. If…that… or Implies4. If and only if or equivalent to5. Negation or Xor

Connectives

1. And the symbol is “ ”

F

F

F

T

P Q

FF

TF

FT

TT

QP

Connectives1. 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

Connectives1. 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

Then we can write be symbol is p q 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

Connectives2. 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

Connectives2. 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

Connectives3. 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

Connectives3. 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

Connectives4. 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

Connectives4. 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

Connectives5. 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 pThus Negation of this statement has truth – value is false

Answer False

Connectives5. 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 pThus 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 ThailandSolve

Give p is Cheangmai used to capital of ThailandGive q is Thonburi used to capital of Thailand

Then we can write be symbol is p qSince p has the truth – value is false

q has the truth – value is trueThus this statement has truth – value is false

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

Solvea b has the truth – value is true and c is falseThus ( a b ) c has truth – value is true

Answer True


Example3) 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

Truth – value of StatementExample4) 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 TF T T

T

T

Answer True

Order to find the Truth – value of Statement

1. Braces2. 3. “” or “”4. 5.

Truth TableYou 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 truth – value be 4 cases * If there is three statements (p, q ,r) then the cases

of the truth – value be 8 cases * If there is four statements (p,q,r,s) then the cases of

the truth – value be 16 cases

Thus if there is n statements then the cases of the truth – value be 2n cases

Truth TableExample.1) Give p and q be statements . Make the truth table

of (p q) (p q)Solve

p q Pq p q pq (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 TableExample.2) Give p and q be statements . Make the truth table

of [(p q) (p q)]Solve

p q Pq q pq (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 TableExample.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 TableExample.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 patternObserve the following examples.1) Give p and q be statements . Make the truth table

of p q and q pSolve

p q p q (p q) q pT 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 pis the same case by case .Then we can say p q equivalent q pThus we can write be symbol is p q q p


of (p q) and p qSolve

p q p q p q (pq) P qT 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 qis the same case by case .Then we can say (p q) equivalent p qThus we can write be symbol is (p q) p q


of p q and p qSolve

p q p p q P qT 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 qis not the same case by case .Then we can say p q is not equivalent p q

Equivalent patternObserve the following examples.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 qT 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 qThus we can write be symbol is (p q) (q p) p q

Equivalent patternObserve the following examples.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”

SolveGive p be 82 is even number

q be 8 is even numberThen 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 patternSymbol of equivalent pattern is “ ”The following two statements are equivalent.1) (p) p2) p q q p3) p q q p4) p q q p5) p (q r) (p q) r6) p (q r) (p q) r7) p (q r) (p q) (p r)8) p (q r) (p q) (p r)9) (p q) p q10) (p q) p q

Equivalent patternSymbol of equivalent pattern is “ ”The following two statements are equivalent.11) (p q) p q12) p q p q13) p q q p 14) p q p q15) p q (p q) (q p)16) p p p17) p p p18) (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 patternSymbol of equivalent pattern is “ ”The following two statements are equivalent.22) p T p

T q q23) p F p

F q q24) T q q p F p25) T q q

p T p F q q p F p

TautologyDefiniton.

The truth – value of statement pattern is true all cases

is called Tautology.

TautologyExample.1) Consider the truth – value of statement pattern [(pq) p] q

p q p q (p q) p [(p q) p] qT T T T T

T F F F T

F T T F T

F F T F T

You can observe [(pq) p] q has the truth – value is true all cases. Then we can say [(pq) p] q is Tautology.

TautologyExample.2) Consider the truth – value of statement pattern (pq) (q p) be Tautology or not.

p q p q p q q p (pq) (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 (pq) (q p) has the truth – value is true all cases. Then we can say (pq) (q p) is Tautology.

TautologyExample.3) Consider the truth – value of statement pattern [(pq) p] q be Tautology or not.

p q p q p q (pq) p [(pq) 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 [(pq) p] q has the truth – value is true not all cases. Then we can say (pq) (q p) is not Tautology.

TautologyBy 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.

TautologyExample.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.


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


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.

TautologyExample.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

ArgumentArgument 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

ArgumentExample.1) Give p and q be statement. Consider the following argument is valid or invalid.

cause 1. p q2. p

conclusion qSolve.

[(p q) p ] qF

T F T T T T

Since [(p q) p ] q is Tautology Thus this argument is valid

Contradiction

ArgumentExample.2) Give p , q and r be statement. Consider the following argument is valid or invalid.

cause 1. p q2. q r3. r

conclusion pSolve.

[(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

ArgumentExample.3) Consider the following argument is valid or invalid.

cause 1. If it rained then the roof is wet2. the roof is not wet

conclusion it is not rainedSolve.

Give p be it rained q be the roof is wet

Then write the symbol.cause 1. p q

2. qconclusion p

Thus we can write [(p q) q ] p

Argument

[(p q) q ] p FT F

T T T T T F

Since [(p q) p ] q is Tautology

Thus this argument is valid

Contradiction

ArgumentExample.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

debtconclusion 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 33. x2 - 2x + 8 = (x – 4)(x + 2)4. x.x = x + x5. 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 x3) U be Universe4) R be Set of real number5) Q be Set of rational number6) I or Z be Set of integer number7) N be Set of natural number

Quantifier

Example. Write the sentence be symbol.

1) For all of x that x + x = 2xAnswer. x[x + x = 2x] ,U = R

2) there are x that x + 0 = 2xAnswer. x[x + 0 = 2x] ,U = R

3) For all of x if x is integer number then x is real numberAnswer. x[x I xR] ,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 1Answer. 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 = RAnswer. 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) yx [x2 + y2 = 8] Answer. For all of y and there are x that x2 + y2 = 8

Quantifier

Example. Write the symbol besentence.

4) yx [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. Quantifier2. Open sentence3. 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 < 5P(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.



2) x[x < 5] , U = I

solve. Give P(x) = x < 5then P(8) = 8 < 5 (False)

Then instead some of x in U to P(x) is falsesome case.Thus x[x < 5] , U = I has truth – value is False.



3) x[x < 5] , U = I

solve. Give P(x) = x < 5then 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 }

solve. Give P(x) = x < 5P(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 sentencep 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

xy[P(x,y)] xy[Px,y)] xy[P(x,y)] xy[P(x,y)]yx[P(x,y)] yx[Px,y)] yx[P(x,y)] yx[P(x,y)]


Definition.

xy[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.

xy[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.

xy[P(x,y)] or yx [P(x,y)]

Example.1) Give U = { -1 , 0 , 1 } Find the truth – value following

Problems.

1.1 xy[xy < 2] Solve.

Give P(x,y) be xy < 2Then 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 xy[xy < 2] is true.


Example.1) Give U = { -1 , 0 , 1 } Find the truth – value

following Problems.

1.2 xy[x + y < 2] Solve.

Give P(x,y) be x + y < 2Then you can see P(1, 1) be 1 + 1 < 2 (False)

Thus xy[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 xy[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 xy[x + y = y + x] is true


Definition.

xy[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.

xy[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.

xy[Px,y)] or yx[Px,y)]


following Problems.

1.1 xy[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 xy[Px,y)] is true.

xy[Px,y)] or yx[Px,y)]


Problems.

1.2 xy[x + y > 2] Solve.

Give P(x,y) be x + y > 2Then 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 xy[x + y > 2] is False.


Definition.

xy[P(x,y)] has the truth – value is true iff instead x with all in universe then y[P(x,y)] is

true.

xy[P(x,y)] has the truth – value is false iff instead x with a some in universe then

y[P(x,y)] is false.

xy[P(x,y)] or yx[P(x,y)]


Problems.

1.1 xy[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 xy[x + y = 0] is true.



following Problems.

1.2 xy[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 xy[x < y] is false.


Example.2) Give U is set of real number .Find the truth – value

following Problems.

2.1 xy[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 xy[x + y = 1] is true.


Example.2) Give U is set of real number .Find the truth – value

following Problems.

2.2 xy[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 xy[x = y2] is false.


Definition.

xy[P(x,y)] has the truth – value is true iff instead x with a some in universe then

y[P(x,y)] is true.

xy[P(x,y)] has the truth – value is false iff instead x with a all in universe then y[P(x,y)]

is false.



following Problems.

1.1 xy [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 xy [x y] is true.



following Problems.

1.2 xy [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 xy [x + y = 0] is false


Example.2) Give U is set of real number. Find the truth – value following Problems.

2.1 xy [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 xy [x + y = y] is true.



2.2 xy [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 xy [x y] is false.

Repeat Lesson.

Example.1) Give U is set of natural number. Find the truth – value following Problems.

1.1 xy [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 xy [y = x + 1] is true.

Repeat Lesson.

Example.1) Give U is set of natural number. Find the truth – value following Problems.

1.2 yx [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 yx [y = x + 1] is false.

Repeat Lesson.


2.1 xy [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 xy [x + y = 0] is true.

Repeat Lesson.


2.2 yx [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 yx [x + y = 0] is false.

Good Luck

Easy Mathematic ^.^

The logic

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