Question

1)How to show an axiomatic system is a model? Please, help
2) How to show 2 axiomatic systems are isomorphic? Please, help

Answer 1

I have an axiomatic system :
Axiom 1: \(if ~~a, b\in \mathbb Z ~~and~~ a < b, then~~ a\neq b\)
Axiom 2: \(if~~ a, b, c\in \mathbb Z, ~~a< b~~and~~b<c, ~~then~~a<c\)
How to show it is a model?

Answer 2

For part 2) I have 2 axiomatic system:
A) \(P\leftrightarrow Bob\)
\(Q\leftrightarrow Ted\)
\(R\leftrightarrow Carol\)
\(\{P, Q\}\leftrightarrow Entertainment ~~committee\)
\(\{P, R\}\leftrightarrow Refreshment ~~committee\)
\(\{Q, R\}\leftrightarrow Finance ~~committee\)

Answer 3

B) \(\{Bob, Ted\}\leftrightarrow P\)
\(\{ Ted, Carol\}\leftrightarrow Q\)
\(\{Bob, Carol \}\leftrightarrow R\)
\(\{P,Q\}\leftrightarrow Bob\)
\(\{P, Q\}\leftrightarrow Ted\)
\(\{Q,R\}\leftrightarrow Carol\)

Answer 4

I don't know how to show that they are isomorphic.

Answer 5

a model is a realization of an axiomatic system; all you've given in 1 is the axioms, not any possible model to verify

Answer 6

By definition of a model, if all axioms are correct, then it is a model. But it means I just say that sentence as a proof?

Answer 7

please post the original questions

Answer 8

"if all axioms are correct" doesn't make sense

Answer 9

That is original one. ok, let me take a pic

Answer 10

Problem 1.2.16

Answer 11

This is the definition of models

Answer 12

okay, so we have a system speaking about \(S,R\) with axioms: $$\forall a,b\in S\ (aRb\implies a\ne b)\\\forall a,b,c\in S\ (aRb\land bRc\implies aRc)$$
then they're asking to verify that the above axioms are valid in the model \((S,R)=(\mathbb Z,<)\), i.e. confirm: $$a<b\implies a\ne b\\a<b\land b<c\implies a<c$$

and likewise for \((\mathbb Z,>)\)

Answer 13

Got you, how about part 2)?

Answer 14

How to show they are isomorphic?

Answer 15

if the models are isomorphic then that means there is a one to one mapping between \(a<b\) i.e. \((a,b)\in\,<\) as a relation and \((a,b)\in\,>\)

Answer 16

and this is true -- we can take \((a,b)\in\, <\) and assign it to \((b,a)\, >\) so \(a<b\Leftrightarrow b>a\) and then our axioms clearly still hold: $$a<b\land b<c\implies a<c\\b>a\land c>b\Leftrightarrow c>b\land b>a\implies c>a$$

Answer 17

and similarly for the first axiom $$a<b\implies a\ne b\\b>a\implies b\ne a\Leftrightarrow a\ne b$$

Answer 18

So, we just show that they are " work" on the same way, right?

Answer 19

@oldrin.bataku Please, explain me part d) I was doubt myself when I think they are isomorphic. I don't know why, just intuitively feel that. I thought in Real, like fraction, we have 1/2 = 3/6 . It is not like what the axiom1 say but don't know how to argue.

Answer 20

in part (d) it is a valid model, but it is not isomorphic because there are uncountably many reals