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$F = \forall x R(x,x)$

$G = \forall x \forall y \forall z (R(x,y) \land R(y,z) \rightarrow R(x,z))$

a) Find a model for $G$, that isn't a model for $F$.

b) Does a domain $D$ exist, so that $val_{D,I,\beta}(F) = true$ for every Interpretation $(D',I)$ with $D = D'$.

c)Does a domain $D$ exist, so that $val_{D,I,\beta}(G) = true$ for every Interpretation $(D',I)$ with $D = D'$.

d)Is $F$ valid? Is $G$ valid? Explain your answer.

I'm doing exercises from a textbook, but there aren't any answers.

a)$D = \mathbb{N}$

$I(R) = \{(x,y) \in D \times D | x < y\}$

The relation is transitive, but can't be reflexive.

I don't understand what is meant in b) and c).

d)For $F$ or $G$ to be valid, they would have to be true for every Interpretation $I$. Like in a) we found an interpretation for which $F$ isn't valid.

Wouldn't this then answer b), that such a domain doesn't exist for every Interpretation $I$.

Is there a simple example for a non transitive relation?

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  • $\begingroup$ Most relations aren't transitive. Just try and build a nontransitive relation on a three-element set. $\endgroup$ – Patrick Stevens Dec 17 '19 at 7:30
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For b) and c), you need to find a set $D$ such that, however you interpret the relation symbol $R$, $F$ (resp. $G$) will hold. Hint : consider $D := \emptyset$.

For d), Just take $D = \{0,1,2\}$ and interpret $R$ as follow : $\{(0,1),(1,2)\}$.

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    $\begingroup$ @franz3 Then the answer for b) is no because you can always interpret $R$ to be the empty relation. For c), the answer is yes : take $D $ to be any singleton. $\endgroup$ – Olivier Roche Dec 17 '19 at 8:44

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