Recursive definition of the relation greater than on N X N

0

$begingroup$

Give a recursive definition of the relation greater than on N X N using the successor operators s?

I started this question throw this way:

basis: (1,0) ∈ N x N

could someone help me in recursive step?

thanks

relations

share|cite|improve this question

asked Sep 21 '14 at 23:23

user3701952user3701952

62

$endgroup$

add a comment | 

0

$begingroup$

Give a recursive definition of the relation greater than on N X N using the successor operators s?

I started this question throw this way:

basis: (1,0) ∈ N x N

could someone help me in recursive step?

thanks

relations

share|cite|improve this question

asked Sep 21 '14 at 23:23

user3701952user3701952

62

$endgroup$

add a comment | 

0

0

0

$begingroup$

Give a recursive definition of the relation greater than on N X N using the successor operators s?

I started this question throw this way:

basis: (1,0) ∈ N x N

could someone help me in recursive step?

thanks

relations

share|cite|improve this question

asked Sep 21 '14 at 23:23

user3701952user3701952

62

$endgroup$

Give a recursive definition of the relation greater than on N X N using the successor operators s?

I started this question throw this way:

basis: (1,0) ∈ N x N

could someone help me in recursive step?

thanks

relations

relations

share|cite|improve this question

asked Sep 21 '14 at 23:23

user3701952user3701952

62

share|cite|improve this question

asked Sep 21 '14 at 23:23

user3701952user3701952

62

share|cite|improve this question

share|cite|improve this question

asked Sep 21 '14 at 23:23

user3701952user3701952

62

asked Sep 21 '14 at 23:23

user3701952user3701952

62

asked Sep 21 '14 at 23:23

user3701952user3701952

62

62

add a comment | 

add a comment | 

4 Answers
4

active

oldest

votes

1

$begingroup$

If $(n,m) in >$ then

$(S(n),m) in >$ and

If $n neq S(m)$

$(n,S(m)) in >$

share|cite|improve this answer

answered Sep 21 '14 at 23:29

Jonas GomesJonas Gomes

1,9711823

$endgroup$

add a comment | 

1

$begingroup$

My favorite recursive definition of $>$ for natural numbers is:

$(0,n) notin ,>$ and

$(m+1,n)in,>iff [(m,n)in,>text{ or }m=n]$.

share|cite|improve this answer

answered Sep 21 '14 at 23:54

Andreas BlassAndreas Blass

50.2k452109

$endgroup$

add a comment | 

0

$begingroup$

How about: if $(a,b) in >$ then $(a+1,b)in >$ and $(a+1,b+1)in >$. (Here I am using $(x,y)in >$ to signify $x>y$.)

share|cite|improve this answer

answered Sep 21 '14 at 23:30

paw88789paw88789

29.4k12349

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice! Simpler than mine!
  $endgroup$
  – Jonas Gomes
  Sep 21 '14 at 23:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jonas Gomes: Yours looks good too. They're pretty similar.
  $endgroup$
  – paw88789
  Sep 21 '14 at 23:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what about the Closure ? [a,b] ∈ N x N it can be obtained from (1,0)
  $endgroup$
  – user3701952
  Sep 21 '14 at 23:51
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

How about:

$$forall x forall y (x < y leftrightarrow (s(x)=y lor exists z (x < z land s(z) = y)))$$

share|cite|improve this answer

answered Jan 22 at 19:45

Bram28Bram28

63.3k44793

$endgroup$

add a comment | 

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

If $(n,m) in >$ then

$(S(n),m) in >$ and

If $n neq S(m)$

$(n,S(m)) in >$

share|cite|improve this answer

answered Sep 21 '14 at 23:29

Jonas GomesJonas Gomes

1,9711823

$endgroup$

add a comment | 

1

$begingroup$

If $(n,m) in >$ then

$(S(n),m) in >$ and

If $n neq S(m)$

$(n,S(m)) in >$

share|cite|improve this answer

answered Sep 21 '14 at 23:29

Jonas GomesJonas Gomes

1,9711823

$endgroup$

add a comment | 

1

1

1

$begingroup$

If $(n,m) in >$ then

$(S(n),m) in >$ and

If $n neq S(m)$

$(n,S(m)) in >$

share|cite|improve this answer

answered Sep 21 '14 at 23:29

Jonas GomesJonas Gomes

1,9711823

$endgroup$

If $(n,m) in >$ then

$(S(n),m) in >$ and

If $n neq S(m)$

$(n,S(m)) in >$

share|cite|improve this answer

answered Sep 21 '14 at 23:29

Jonas GomesJonas Gomes

1,9711823

share|cite|improve this answer

share|cite|improve this answer

answered Sep 21 '14 at 23:29

Jonas GomesJonas Gomes

1,9711823

answered Sep 21 '14 at 23:29

Jonas GomesJonas Gomes

1,9711823

answered Sep 21 '14 at 23:29

Jonas GomesJonas Gomes

1,9711823

1,9711823

add a comment | 

add a comment | 

1

$begingroup$

My favorite recursive definition of $>$ for natural numbers is:

$(0,n) notin ,>$ and

$(m+1,n)in,>iff [(m,n)in,>text{ or }m=n]$.

share|cite|improve this answer

answered Sep 21 '14 at 23:54

Andreas BlassAndreas Blass

50.2k452109

$endgroup$

add a comment | 

1

$begingroup$

My favorite recursive definition of $>$ for natural numbers is:

$(0,n) notin ,>$ and

$(m+1,n)in,>iff [(m,n)in,>text{ or }m=n]$.

share|cite|improve this answer

answered Sep 21 '14 at 23:54

Andreas BlassAndreas Blass

50.2k452109

$endgroup$

add a comment | 

1

1

1

$begingroup$

My favorite recursive definition of $>$ for natural numbers is:

$(0,n) notin ,>$ and

$(m+1,n)in,>iff [(m,n)in,>text{ or }m=n]$.

share|cite|improve this answer

answered Sep 21 '14 at 23:54

Andreas BlassAndreas Blass

50.2k452109

$endgroup$

My favorite recursive definition of $>$ for natural numbers is:

$(0,n) notin ,>$ and

$(m+1,n)in,>iff [(m,n)in,>text{ or }m=n]$.

share|cite|improve this answer

answered Sep 21 '14 at 23:54

Andreas BlassAndreas Blass

50.2k452109

share|cite|improve this answer

share|cite|improve this answer

answered Sep 21 '14 at 23:54

Andreas BlassAndreas Blass

50.2k452109

answered Sep 21 '14 at 23:54

Andreas BlassAndreas Blass

50.2k452109

answered Sep 21 '14 at 23:54

Andreas BlassAndreas Blass

50.2k452109

50.2k452109

add a comment | 

add a comment | 

0

$begingroup$

How about: if $(a,b) in >$ then $(a+1,b)in >$ and $(a+1,b+1)in >$. (Here I am using $(x,y)in >$ to signify $x>y$.)

share|cite|improve this answer

answered Sep 21 '14 at 23:30

paw88789paw88789

29.4k12349

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice! Simpler than mine!
  $endgroup$
  – Jonas Gomes
  Sep 21 '14 at 23:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jonas Gomes: Yours looks good too. They're pretty similar.
  $endgroup$
  – paw88789
  Sep 21 '14 at 23:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what about the Closure ? [a,b] ∈ N x N it can be obtained from (1,0)
  $endgroup$
  – user3701952
  Sep 21 '14 at 23:51
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

How about: if $(a,b) in >$ then $(a+1,b)in >$ and $(a+1,b+1)in >$. (Here I am using $(x,y)in >$ to signify $x>y$.)

share|cite|improve this answer

answered Sep 21 '14 at 23:30

paw88789paw88789

29.4k12349

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice! Simpler than mine!
  $endgroup$
  – Jonas Gomes
  Sep 21 '14 at 23:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jonas Gomes: Yours looks good too. They're pretty similar.
  $endgroup$
  – paw88789
  Sep 21 '14 at 23:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what about the Closure ? [a,b] ∈ N x N it can be obtained from (1,0)
  $endgroup$
  – user3701952
  Sep 21 '14 at 23:51
  
  
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

How about: if $(a,b) in >$ then $(a+1,b)in >$ and $(a+1,b+1)in >$. (Here I am using $(x,y)in >$ to signify $x>y$.)

share|cite|improve this answer

answered Sep 21 '14 at 23:30

paw88789paw88789

29.4k12349

$endgroup$

How about: if $(a,b) in >$ then $(a+1,b)in >$ and $(a+1,b+1)in >$. (Here I am using $(x,y)in >$ to signify $x>y$.)

share|cite|improve this answer

answered Sep 21 '14 at 23:30

paw88789paw88789

29.4k12349

share|cite|improve this answer

share|cite|improve this answer

answered Sep 21 '14 at 23:30

paw88789paw88789

29.4k12349

answered Sep 21 '14 at 23:30

paw88789paw88789

29.4k12349

answered Sep 21 '14 at 23:30

paw88789paw88789

29.4k12349

29.4k12349

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice! Simpler than mine!
  $endgroup$
  – Jonas Gomes
  Sep 21 '14 at 23:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jonas Gomes: Yours looks good too. They're pretty similar.
  $endgroup$
  – paw88789
  Sep 21 '14 at 23:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what about the Closure ? [a,b] ∈ N x N it can be obtained from (1,0)
  $endgroup$
  – user3701952
  Sep 21 '14 at 23:51
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice! Simpler than mine!
  $endgroup$
  – Jonas Gomes
  Sep 21 '14 at 23:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jonas Gomes: Yours looks good too. They're pretty similar.
  $endgroup$
  – paw88789
  Sep 21 '14 at 23:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what about the Closure ? [a,b] ∈ N x N it can be obtained from (1,0)
  $endgroup$
  – user3701952
  Sep 21 '14 at 23:51
  
  
  

  
  
  
  

$begingroup$
Nice! Simpler than mine!
$endgroup$
– Jonas Gomes
Sep 21 '14 at 23:32

$begingroup$
Nice! Simpler than mine!
$endgroup$
– Jonas Gomes
Sep 21 '14 at 23:32

$begingroup$
@Jonas Gomes: Yours looks good too. They're pretty similar.
$endgroup$
– paw88789
Sep 21 '14 at 23:34

$begingroup$
@Jonas Gomes: Yours looks good too. They're pretty similar.
$endgroup$
– paw88789
Sep 21 '14 at 23:34

$begingroup$
what about the Closure ? [a,b] ∈ N x N it can be obtained from (1,0)
$endgroup$
– user3701952
Sep 21 '14 at 23:51

$begingroup$
what about the Closure ? [a,b] ∈ N x N it can be obtained from (1,0)
$endgroup$
– user3701952
Sep 21 '14 at 23:51

add a comment | 

0

$begingroup$

How about:

$$forall x forall y (x < y leftrightarrow (s(x)=y lor exists z (x < z land s(z) = y)))$$

share|cite|improve this answer

answered Jan 22 at 19:45

Bram28Bram28

63.3k44793

$endgroup$

add a comment | 

0

$begingroup$

How about:

$$forall x forall y (x < y leftrightarrow (s(x)=y lor exists z (x < z land s(z) = y)))$$

share|cite|improve this answer

answered Jan 22 at 19:45

Bram28Bram28

63.3k44793

$endgroup$

add a comment | 

0

0

0

$begingroup$

How about:

$$forall x forall y (x < y leftrightarrow (s(x)=y lor exists z (x < z land s(z) = y)))$$

share|cite|improve this answer

answered Jan 22 at 19:45

Bram28Bram28

63.3k44793

$endgroup$

How about:

$$forall x forall y (x < y leftrightarrow (s(x)=y lor exists z (x < z land s(z) = y)))$$

share|cite|improve this answer

answered Jan 22 at 19:45

Bram28Bram28

63.3k44793

share|cite|improve this answer

share|cite|improve this answer

answered Jan 22 at 19:45

Bram28Bram28

63.3k44793

answered Jan 22 at 19:45

Bram28Bram28

63.3k44793

answered Jan 22 at 19:45

Bram28Bram28

63.3k44793

63.3k44793

add a comment | 

add a comment | 

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But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

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Use MathJax to format equations. MathJax reference.

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