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Let $\Sigma=\{A,B,C,D,E\}$

$A\equiv \forall x e(x,x)$

$B\equiv \forall x \forall y e(x,y) \to e(y,x)$

$C\equiv \forall x \forall y \forall z( e(x,y) \land e(y,z))\to e(x,z)$

$D\equiv \forall x \forall ye(x,y)\to e(f(x),f(y))$

$E\equiv \forall x \forall y e(x,y) \to (p(x) \leftarrow \rightarrow p(y))$

Where $p$ is a predicate and $f$ is a function.

I have to show that there is no model $M$ such that $T_\Sigma=T_M$.

For this i have to find a formula $A$ such that $T_{\Sigma \cup A}$ and $T_{\Sigma \cup \lnot A}$ are both consistent.

$A$ has to be a closed formula.

$T_\Sigma$ is the theory produced from $\Sigma$

Can i get some help?

asddf
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1 Answers1

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The axioms say $e$ is an equivalence relation, that $f$ and $p$ respect the relation by having $f$ take all the elements in a given class to an image in another or the same class and having $p$ have the same truth value for all elements of the same class. I can think of two statements that can be added. One I can think of is $\forall x p(x)$. This can clearly be added to your axioms without a problem. Its negation is $\exists x \lnot p(x)$. As long as there is at least one element in the universe, we can strengthen this to $\forall x \lnot p(x)$ and this is consistent as well. As long as there are two elements in the universe another is $\exists x,y f(x) \neq f(y)$ or $\forall x,y f(x)=f(y)$ without a problem.

Ross Millikan
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  • I understand the question and the comments and the answer,but I do not understand your statement by OP:I have to show that there is no model $M$ such that $T_\Sigma=T_M$. I can easily find many interpretations (models)) for \Sigma. – magma Oct 08 '17 at 14:49
  • Please see updated the comment – magma Oct 08 '17 at 14:52
  • @magma: as I am used to it, it is incorrect to talk of $T_M$. A theory is the set of consequences of a set of axioms. I have found a sentence that is not decided by the first set of axioms, as shown by the fact that you can add either it or its negation to the axioms and get a consistent theory. That means the theory with the sentence is a superset of the theory without it. – Ross Millikan Oct 08 '17 at 15:00
  • I completely agree with you. Exactly my point. So why does the OP say that there are no models for\Sigma? – magma Oct 08 '17 at 15:26
  • @magma how i understand, there are models for $\Sigma$ just there isnt a model M such that $T _\Sigma = T_M$ – asddf Oct 08 '17 at 17:08
  • By model i mean interpretation, im sorry if that confused you. – asddf Oct 08 '17 at 17:10
  • There are models for $\Sigma$, but each one makes $\forall x p(x)$ either true or false. That shows that $T_{\Sigma}$ is not complete. The best I can guess for $T_M$ is that it is a complete set of sentences true in the model $M$ – Ross Millikan Oct 08 '17 at 19:24