# Logical Implication

#### skamoni

Let $$\displaystyle \theta$$ be $$\displaystyle \forall{x}\forall{y}\exists{z}((R(x,z)\wedge {R(y,z)})\wedge \forall{w}((R(x,w)\wedge {R(y,w)}) \rightarrow R(z,w)))$$ and let $$\displaystyle \phi$$ be $$\displaystyle \forall{x}\forall{y}\exists{z}((R(z,x)\wedge {R(z,y)})\wedge \forall{w}((R(w,x)\wedge {R(w,y)}) \rightarrow R(w,z)))$$

Where R is a binary predicate symbol.

I have been asked to find an L structure consisting of 3 elements such that $$\displaystyle \theta \not \models \phi$$

My Professor hasn't really given us a method to do this, any help would be appreciated.

#### clic-clac

Hi

If I understand, you want a 3 elements structure where $$\displaystyle \theta$$ is true but $$\displaystyle \phi$$ isn't. In these sentences, precising a binary relation symbol properties, you can try to see $$\displaystyle R$$ as an order relation (perhaps the most common binary predicates are order or equivalence relations).
In such context, $$\displaystyle \theta$$ says that for any $$\displaystyle x,y$$, there is a $$\displaystyle \sup(x,y)$$ (for $$\displaystyle R$$), and $$\displaystyle \phi$$ that for any $$\displaystyle x,y$$, there is a $$\displaystyle \inf(x,y)$$.

Then I guess you can quite easily define a partial order on three elements such that any two of them have a sup, but there are two with no inf (and in fact more: with no lower bound).

• skamoni