# Logical Implication

• May 23rd 2010, 04:38 AM
skamoni
Logical Implication
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.
• May 25th 2010, 01:33 AM
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).