# Predicate logic: Finding a domain for every interpretation

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

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

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?

• Most relations aren't transitive. Just try and build a nontransitive relation on a three-element set. – Patrick Stevens Dec 17 '19 at 7:30

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 : $$\{(0,1),(1,2)\}$$.
• @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. – Olivier Roche Dec 17 '19 at 8:44