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I am trying to solve these problems :

(a) Define a $V_G$-sentence $\phi$ such that $\phi$ has arbitrarily large finite models, and for any model $G$, $G$ is a connected graph.

(b) Find a connected graph that does not model the sentence $\phi$ you found in (a).

Any ideas?

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closed as off-topic by TheGeekGreek, Namaste, Stefan Mesken, Davide Giraudo, Leucippus Jun 1 '17 at 0:04

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  • 3
    $\begingroup$ Can you clarify what $V$ is? $\endgroup$ – Hanul Jeon May 1 '17 at 18:08
  • $\begingroup$ $V$ is the vocabulary over $G$, so actually we are looking for a $V_{G}$-sentence. $\endgroup$ – Angela May 1 '17 at 22:48
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The proof that graph connectedness is not expressible in the first-order language of graphs is a classic application of compactness. The purpose of this exercise is to show that a first-order sentence in the language of graphs may only have connected models, as long as it excludes some connected graphs as well.

This observation suggests how to proceed. A sentence about graphs that has connected models of arbitrary size is one that characterizes complete (simple) graphs. If $E$ is the edge relation, then

$$ \forall x \,.\, \neg E(x,x) \wedge \forall y \,.\, x=y \vee E(x,y) \enspace.$$

Every connected but incomplete graph is not a model of this sentence.

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  • $\begingroup$ Thank you for your response. Just one last question to be sure, what this symbol, $.$, mean in your sentence. Is it multiplication? $\endgroup$ – Angela May 30 '17 at 12:59
  • $\begingroup$ No, the dot, borrowed from lambda calculus, separates the quantifier (e.g., $\forall x$) from the expression that is its scope. There's plenty of different notations in use for quantification formulae. An alternative to the one above would be $(\forall x)(\neg E(x,x) \wedge (\forall y)(x=y \vee E(x,y)))$. Once you get used to the dots, it's a less cluttered notation. $\endgroup$ – Fabio Somenzi May 30 '17 at 14:19
  • $\begingroup$ Yes that is true!Thank you so much for your help! :-) $\endgroup$ – Angela May 30 '17 at 14:52

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