Recursive definition of the relation greater than on N X N

Question

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

score 1 · Accepted Answer · answered Sep 21 '14 at 23:29

1

If $(n,m) \in \ >$ then

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

If $n \neq S(m)$

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

answered Sep 21 '14 at 23:29

Jonas Gomes

score 1 · Answer 2 · answered Sep 21 '14 at 23:54

1

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

$(0,n) \notin \,>$ and

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

answered Sep 21 '14 at 23:54

Andreas Blass

score 0 · Answer 3 · answered Jan 22 '19 at 19:45

0

How about:

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

answered Jan 22 '19 at 19:45

Bram28

score 0 · Answer 4 · answered Sep 21 '14 at 23:30

0

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 $x>y$.)

answered Sep 21 '14 at 23:30

paw88789

Nice! Simpler than mine! – Jonas Gomes Sep 21 '14 at 23:32
@Jonas Gomes: Yours looks good too. They're pretty similar. – paw88789 Sep 21 '14 at 23:34
what about the Closure ? [a,b] ∈ N x N it can be obtained from (1,0) – user3701952 Sep 21 '14 at 23:51

4 Answers4