Logical Implication

Apr 2008
50
8
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.
 
Nov 2008
354
185
Paris
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).
 
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