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Loser66
 one year ago
1)How to show an axiomatic system is a model? Please, help
2) How to show 2 axiomatic systems are isomorphic? Please, help
Loser66
 one year ago
1)How to show an axiomatic system is a model? Please, help 2) How to show 2 axiomatic systems are isomorphic? Please, help

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I 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?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1For part 2) I have 2 axiomatic system: A) \(P\leftrightarrow Bob\) \(Q\leftrightarrow Ted\) \(R\leftrightarrow Carol\) \(\{P, Q\}\leftrightarrow Entertainment ~~committee\) \(\{P, R\}\leftrightarrow Refreshment ~~committee\) \(\{Q, R\}\leftrightarrow Finance ~~committee\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1B) \(\{Bob, Ted\}\leftrightarrow P\) \(\{ Ted, Carol\}\leftrightarrow Q\) \(\{Bob, Carol \}\leftrightarrow R\) \(\{P,Q\}\leftrightarrow Bob\) \(\{P, Q\}\leftrightarrow Ted\) \(\{Q,R\}\leftrightarrow Carol\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I don't know how to show that they are isomorphic.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a model is a realization of an axiomatic system; all you've given in 1 is the axioms, not any possible model to verify

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1By 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0please post the original questions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0"if all axioms are correct" doesn't make sense

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1That is original one. ok, let me take a pic

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1This is the definition of models

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, 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 \((\mathbb Z,>)\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Got you, how about part 2)?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1How to show they are isomorphic?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if 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 \((a,b)\in\,>\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and 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: $$a<b\land b<c\implies a<c\\b>a\land c>b\Leftrightarrow c>b\land b>a\implies c>a$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and similarly for the first axiom $$a<b\implies a\ne b\\b>a\implies b\ne a\Leftrightarrow a\ne b$$

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1So, we just show that they are " work" on the same way, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in part (d) it is a valid model, but it is not isomorphic because there are uncountably many reals
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