Which of the following logic statements are true

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$S_1: ,,forall x ,exists y , forall z,(x+y=z)$

$S_2: ,, exists x , forall y , exists z,(x+y=z)$

where $x,y, textand ,,z $ are real numbers.
Which of the following statements are true?

My Approach

I think both $S_1$ and $S_2$ are true.
Because $S_1$ says for all $x$,there exists a $y$ whose sum will be $z$, a real number as sum of two real numbers are real.hence $S_1$ holds

same reasoning can be given for $S_2$ .but the answer says $S_2$ only.I know i am making some mistake ,don't know where.please help

logic first-order-logic predicate-logic

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edited Aug 9 at 17:46

Taroccoesbrocco

3,48941431

asked Jul 20 at 6:03

laura

1,238521

add a comment | 

up vote
3
down vote

favorite

Question

$S_1: ,,forall x ,exists y , forall z,(x+y=z)$

$S_2: ,, exists x , forall y , exists z,(x+y=z)$

where $x,y, textand ,,z $ are real numbers.
Which of the following statements are true?

My Approach

I think both $S_1$ and $S_2$ are true.
Because $S_1$ says for all $x$,there exists a $y$ whose sum will be $z$, a real number as sum of two real numbers are real.hence $S_1$ holds

same reasoning can be given for $S_2$ .but the answer says $S_2$ only.I know i am making some mistake ,don't know where.please help

logic first-order-logic predicate-logic

share|cite|improve this question

edited Aug 9 at 17:46

Taroccoesbrocco

3,48941431

asked Jul 20 at 6:03

laura

1,238521

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

Question

$S_1: ,,forall x ,exists y , forall z,(x+y=z)$

$S_2: ,, exists x , forall y , exists z,(x+y=z)$

where $x,y, textand ,,z $ are real numbers.
Which of the following statements are true?

My Approach

I think both $S_1$ and $S_2$ are true.
Because $S_1$ says for all $x$,there exists a $y$ whose sum will be $z$, a real number as sum of two real numbers are real.hence $S_1$ holds

same reasoning can be given for $S_2$ .but the answer says $S_2$ only.I know i am making some mistake ,don't know where.please help

logic first-order-logic predicate-logic

share|cite|improve this question

edited Aug 9 at 17:46

Taroccoesbrocco

3,48941431

asked Jul 20 at 6:03

laura

1,238521

Question

$S_1: ,,forall x ,exists y , forall z,(x+y=z)$

$S_2: ,, exists x , forall y , exists z,(x+y=z)$

where $x,y, textand ,,z $ are real numbers.
Which of the following statements are true?

My Approach

I think both $S_1$ and $S_2$ are true.
Because $S_1$ says for all $x$,there exists a $y$ whose sum will be $z$, a real number as sum of two real numbers are real.hence $S_1$ holds

same reasoning can be given for $S_2$ .but the answer says $S_2$ only.I know i am making some mistake ,don't know where.please help

logic first-order-logic predicate-logic

share|cite|improve this question

edited Aug 9 at 17:46

Taroccoesbrocco

3,48941431

asked Jul 20 at 6:03

laura

1,238521

share|cite|improve this question

share|cite|improve this question

edited Aug 9 at 17:46

Taroccoesbrocco

3,48941431

edited Aug 9 at 17:46

Taroccoesbrocco

3,48941431

edited Aug 9 at 17:46

Taroccoesbrocco

3,48941431

3,48941431

asked Jul 20 at 6:03

laura

1,238521

asked Jul 20 at 6:03

laura

1,238521

asked Jul 20 at 6:03

laura

1,238521

1,238521

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote



accepted

$S_1$ is false. Its negation is
$$exists xforall yexists z x+yne z$$
and this is trivially true: take $x=0$, then for any $y$ take $z=y+1$.

$S_2$ is true, since given $x=0$ and some $y$ we can take $z=y$ to make the statement true.

share|cite|improve this answer

answered Jul 20 at 6:06

Parcly Taxel

33.6k136588

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Negation of $S_2$ is $forall x exists y forall z (x+y neq z)$ which should be false if we take $y=0$, then for every $x$ ,$y$ ,$x=y$ Right?
  – laura
  Jul 20 at 6:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @:parcly i wrote negation of $S_2$ .for $S_2$ to be true ,its negation should be false ...right?
  – laura
  Jul 20 at 6:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @laura OK, $neg S_2$ is false, since when $x=0$, there is no $y$ such that for all $z$ $x+y=z$ or $y=z$. Since $x=0$ fails the outer $forall$, the whole negation is false.
  – Parcly Taxel
  Jul 20 at 6:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks ..that is what u wa asking in my first comment...!
  – laura
  Jul 20 at 6:27
  

  
  
  
  

add a comment | 

up vote
1
down vote

Because $S_1$ says for all $x$, there exists a $y$ whose sum will be $z$, a real number as sum of two real numbers are real.

No, $forall x ,exists y , forall z,(x+y=z)$ sais that for any $x$, there is a $y$, such that every $z$ will equal their sum ($x+y$).

Whereas $exists x , forall y , exists z,(x+y=z)$ sais there is an $x$, such that for every $y$, there is a $z$ equal to their sum ($x+y$).

share|cite|improve this answer

answered Jul 20 at 6:53

Graham Kemp

80.1k43275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @:graham sir , thanks for your response. Using negation , i am able to understand it.but answering it straight away with your explanation is little confusing for me . $S_1$ says for any $x,$ there is a $y$, such that every $z$ will equal their sum $(x+y)$ here too for any $x$,if i take $y=0$, then sum $z$ is equal to $x+0=x$. why it is false ? please help me out!
  – laura
  Jul 20 at 7:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Every $z$, not just some $z$. All of them. Take any $x$ and some $y$, can $x+y$ equal every $z$? @laura
  – Graham Kemp
  Jul 20 at 7:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes ..i got it ..thanks a lot!
  – laura
  Jul 20 at 7:57
  

  
  
  
  

add a comment | 

Your Answer

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

$S_1$ is false. Its negation is
$$exists xforall yexists z x+yne z$$
and this is trivially true: take $x=0$, then for any $y$ take $z=y+1$.

$S_2$ is true, since given $x=0$ and some $y$ we can take $z=y$ to make the statement true.

share|cite|improve this answer

answered Jul 20 at 6:06

Parcly Taxel

33.6k136588

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Negation of $S_2$ is $forall x exists y forall z (x+y neq z)$ which should be false if we take $y=0$, then for every $x$ ,$y$ ,$x=y$ Right?
  – laura
  Jul 20 at 6:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @:parcly i wrote negation of $S_2$ .for $S_2$ to be true ,its negation should be false ...right?
  – laura
  Jul 20 at 6:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @laura OK, $neg S_2$ is false, since when $x=0$, there is no $y$ such that for all $z$ $x+y=z$ or $y=z$. Since $x=0$ fails the outer $forall$, the whole negation is false.
  – Parcly Taxel
  Jul 20 at 6:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks ..that is what u wa asking in my first comment...!
  – laura
  Jul 20 at 6:27
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

$S_1$ is false. Its negation is
$$exists xforall yexists z x+yne z$$
and this is trivially true: take $x=0$, then for any $y$ take $z=y+1$.

$S_2$ is true, since given $x=0$ and some $y$ we can take $z=y$ to make the statement true.

share|cite|improve this answer

answered Jul 20 at 6:06

Parcly Taxel

33.6k136588

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Negation of $S_2$ is $forall x exists y forall z (x+y neq z)$ which should be false if we take $y=0$, then for every $x$ ,$y$ ,$x=y$ Right?
  – laura
  Jul 20 at 6:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @:parcly i wrote negation of $S_2$ .for $S_2$ to be true ,its negation should be false ...right?
  – laura
  Jul 20 at 6:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @laura OK, $neg S_2$ is false, since when $x=0$, there is no $y$ such that for all $z$ $x+y=z$ or $y=z$. Since $x=0$ fails the outer $forall$, the whole negation is false.
  – Parcly Taxel
  Jul 20 at 6:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks ..that is what u wa asking in my first comment...!
  – laura
  Jul 20 at 6:27
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

$S_1$ is false. Its negation is
$$exists xforall yexists z x+yne z$$
and this is trivially true: take $x=0$, then for any $y$ take $z=y+1$.

$S_2$ is true, since given $x=0$ and some $y$ we can take $z=y$ to make the statement true.

share|cite|improve this answer

answered Jul 20 at 6:06

Parcly Taxel

33.6k136588

$S_1$ is false. Its negation is
$$exists xforall yexists z x+yne z$$
and this is trivially true: take $x=0$, then for any $y$ take $z=y+1$.

$S_2$ is true, since given $x=0$ and some $y$ we can take $z=y$ to make the statement true.

share|cite|improve this answer

answered Jul 20 at 6:06

Parcly Taxel

33.6k136588

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 6:06

Parcly Taxel

33.6k136588

answered Jul 20 at 6:06

Parcly Taxel

33.6k136588

answered Jul 20 at 6:06

Parcly Taxel

33.6k136588

33.6k136588

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Negation of $S_2$ is $forall x exists y forall z (x+y neq z)$ which should be false if we take $y=0$, then for every $x$ ,$y$ ,$x=y$ Right?
  – laura
  Jul 20 at 6:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @:parcly i wrote negation of $S_2$ .for $S_2$ to be true ,its negation should be false ...right?
  – laura
  Jul 20 at 6:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @laura OK, $neg S_2$ is false, since when $x=0$, there is no $y$ such that for all $z$ $x+y=z$ or $y=z$. Since $x=0$ fails the outer $forall$, the whole negation is false.
  – Parcly Taxel
  Jul 20 at 6:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks ..that is what u wa asking in my first comment...!
  – laura
  Jul 20 at 6:27
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Negation of $S_2$ is $forall x exists y forall z (x+y neq z)$ which should be false if we take $y=0$, then for every $x$ ,$y$ ,$x=y$ Right?
  – laura
  Jul 20 at 6:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @:parcly i wrote negation of $S_2$ .for $S_2$ to be true ,its negation should be false ...right?
  – laura
  Jul 20 at 6:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @laura OK, $neg S_2$ is false, since when $x=0$, there is no $y$ such that for all $z$ $x+y=z$ or $y=z$. Since $x=0$ fails the outer $forall$, the whole negation is false.
  – Parcly Taxel
  Jul 20 at 6:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks ..that is what u wa asking in my first comment...!
  – laura
  Jul 20 at 6:27
  

  
  
  
  

Negation of $S_2$ is $forall x exists y forall z (x+y neq z)$ which should be false if we take $y=0$, then for every $x$ ,$y$ ,$x=y$ Right?
– laura
Jul 20 at 6:21

Negation of $S_2$ is $forall x exists y forall z (x+y neq z)$ which should be false if we take $y=0$, then for every $x$ ,$y$ ,$x=y$ Right?
– laura
Jul 20 at 6:21

@:parcly i wrote negation of $S_2$ .for $S_2$ to be true ,its negation should be false ...right?
– laura
Jul 20 at 6:24

@:parcly i wrote negation of $S_2$ .for $S_2$ to be true ,its negation should be false ...right?
– laura
Jul 20 at 6:24

@laura OK, $neg S_2$ is false, since when $x=0$, there is no $y$ such that for all $z$ $x+y=z$ or $y=z$. Since $x=0$ fails the outer $forall$, the whole negation is false.
– Parcly Taxel
Jul 20 at 6:25

@laura OK, $neg S_2$ is false, since when $x=0$, there is no $y$ such that for all $z$ $x+y=z$ or $y=z$. Since $x=0$ fails the outer $forall$, the whole negation is false.
– Parcly Taxel
Jul 20 at 6:25

thanks ..that is what u wa asking in my first comment...!
– laura
Jul 20 at 6:27

thanks ..that is what u wa asking in my first comment...!
– laura
Jul 20 at 6:27

add a comment | 

up vote
1
down vote

Because $S_1$ says for all $x$, there exists a $y$ whose sum will be $z$, a real number as sum of two real numbers are real.

No, $forall x ,exists y , forall z,(x+y=z)$ sais that for any $x$, there is a $y$, such that every $z$ will equal their sum ($x+y$).

Whereas $exists x , forall y , exists z,(x+y=z)$ sais there is an $x$, such that for every $y$, there is a $z$ equal to their sum ($x+y$).

share|cite|improve this answer

answered Jul 20 at 6:53

Graham Kemp

80.1k43275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @:graham sir , thanks for your response. Using negation , i am able to understand it.but answering it straight away with your explanation is little confusing for me . $S_1$ says for any $x,$ there is a $y$, such that every $z$ will equal their sum $(x+y)$ here too for any $x$,if i take $y=0$, then sum $z$ is equal to $x+0=x$. why it is false ? please help me out!
  – laura
  Jul 20 at 7:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Every $z$, not just some $z$. All of them. Take any $x$ and some $y$, can $x+y$ equal every $z$? @laura
  – Graham Kemp
  Jul 20 at 7:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes ..i got it ..thanks a lot!
  – laura
  Jul 20 at 7:57
  

  
  
  
  

add a comment | 

up vote
1
down vote

Because $S_1$ says for all $x$, there exists a $y$ whose sum will be $z$, a real number as sum of two real numbers are real.

No, $forall x ,exists y , forall z,(x+y=z)$ sais that for any $x$, there is a $y$, such that every $z$ will equal their sum ($x+y$).

Whereas $exists x , forall y , exists z,(x+y=z)$ sais there is an $x$, such that for every $y$, there is a $z$ equal to their sum ($x+y$).

share|cite|improve this answer

answered Jul 20 at 6:53

Graham Kemp

80.1k43275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @:graham sir , thanks for your response. Using negation , i am able to understand it.but answering it straight away with your explanation is little confusing for me . $S_1$ says for any $x,$ there is a $y$, such that every $z$ will equal their sum $(x+y)$ here too for any $x$,if i take $y=0$, then sum $z$ is equal to $x+0=x$. why it is false ? please help me out!
  – laura
  Jul 20 at 7:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Every $z$, not just some $z$. All of them. Take any $x$ and some $y$, can $x+y$ equal every $z$? @laura
  – Graham Kemp
  Jul 20 at 7:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes ..i got it ..thanks a lot!
  – laura
  Jul 20 at 7:57
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Because $S_1$ says for all $x$, there exists a $y$ whose sum will be $z$, a real number as sum of two real numbers are real.

No, $forall x ,exists y , forall z,(x+y=z)$ sais that for any $x$, there is a $y$, such that every $z$ will equal their sum ($x+y$).

Whereas $exists x , forall y , exists z,(x+y=z)$ sais there is an $x$, such that for every $y$, there is a $z$ equal to their sum ($x+y$).

share|cite|improve this answer

answered Jul 20 at 6:53

Graham Kemp

80.1k43275

Because $S_1$ says for all $x$, there exists a $y$ whose sum will be $z$, a real number as sum of two real numbers are real.

No, $forall x ,exists y , forall z,(x+y=z)$ sais that for any $x$, there is a $y$, such that every $z$ will equal their sum ($x+y$).

Whereas $exists x , forall y , exists z,(x+y=z)$ sais there is an $x$, such that for every $y$, there is a $z$ equal to their sum ($x+y$).

share|cite|improve this answer

answered Jul 20 at 6:53

Graham Kemp

80.1k43275

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 6:53

Graham Kemp

80.1k43275

answered Jul 20 at 6:53

Graham Kemp

80.1k43275

answered Jul 20 at 6:53

Graham Kemp

80.1k43275

80.1k43275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @:graham sir , thanks for your response. Using negation , i am able to understand it.but answering it straight away with your explanation is little confusing for me . $S_1$ says for any $x,$ there is a $y$, such that every $z$ will equal their sum $(x+y)$ here too for any $x$,if i take $y=0$, then sum $z$ is equal to $x+0=x$. why it is false ? please help me out!
  – laura
  Jul 20 at 7:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Every $z$, not just some $z$. All of them. Take any $x$ and some $y$, can $x+y$ equal every $z$? @laura
  – Graham Kemp
  Jul 20 at 7:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes ..i got it ..thanks a lot!
  – laura
  Jul 20 at 7:57
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @:graham sir , thanks for your response. Using negation , i am able to understand it.but answering it straight away with your explanation is little confusing for me . $S_1$ says for any $x,$ there is a $y$, such that every $z$ will equal their sum $(x+y)$ here too for any $x$,if i take $y=0$, then sum $z$ is equal to $x+0=x$. why it is false ? please help me out!
  – laura
  Jul 20 at 7:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Every $z$, not just some $z$. All of them. Take any $x$ and some $y$, can $x+y$ equal every $z$? @laura
  – Graham Kemp
  Jul 20 at 7:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes ..i got it ..thanks a lot!
  – laura
  Jul 20 at 7:57
  

  
  
  
  

@:graham sir , thanks for your response. Using negation , i am able to understand it.but answering it straight away with your explanation is little confusing for me . $S_1$ says for any $x,$ there is a $y$, such that every $z$ will equal their sum $(x+y)$ here too for any $x$,if i take $y=0$, then sum $z$ is equal to $x+0=x$. why it is false ? please help me out!
– laura
Jul 20 at 7:19

@:graham sir , thanks for your response. Using negation , i am able to understand it.but answering it straight away with your explanation is little confusing for me . $S_1$ says for any $x,$ there is a $y$, such that every $z$ will equal their sum $(x+y)$ here too for any $x$,if i take $y=0$, then sum $z$ is equal to $x+0=x$. why it is false ? please help me out!
– laura
Jul 20 at 7:19

1

1

Every $z$, not just some $z$. All of them. Take any $x$ and some $y$, can $x+y$ equal every $z$? @laura
– Graham Kemp
Jul 20 at 7:56

Every $z$, not just some $z$. All of them. Take any $x$ and some $y$, can $x+y$ equal every $z$? @laura
– Graham Kemp
Jul 20 at 7:56

yes ..i got it ..thanks a lot!
– laura
Jul 20 at 7:57

yes ..i got it ..thanks a lot!
– laura
Jul 20 at 7:57

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