If I have the following formula F and want to proof that a Model fulfills F.
$$F = \forall a \exists b \exists c [(P(a,b) \land P(c,b) \land P(a,c)) \implies P(c,a) ]$$
I have the following Model $M = <\mathbb{N}, {(g,h)|g < h}, >$, where $\mathbb{N}$ contains all positive integers and also $0$, but no negative integers.
So, for Interpretation I(P) = g < h, we get the following:
$$a < b \land c < b \land a<c \implies c < a$$
How do I get further from this point? How can I proof that the Modell M fulfills or does not fufill F?
Thanks in advance!
You have to show that $M \vDash F$, i.e. that :
We can consider two cases :
i) $a=0$.
In this case $(c < 0)$ does not hold for $c$ whatever, i.e. there is no $c$ such that $c < 0$.
Thus, the conditional has a False consequent and so, in order to be satisfied, also the antecedent must be False.
But for $a=0$ the conditional amounts to : $(0<b ∧ c<b ∧ 0<c \to c<0)$ and it is enough to choose again the value $0$ for both $b$ and $c$ and we have that : $(0<0 ∧ 0<0 ∧ 0<0 \to 0<0)$ is True.
ii) $a > 0$.
In this case we have that : $(c<a)$ is satisfied for $c=0$, i.e. there is a $c$ such that $c < a$.
Thus, the conditional has a True consequent and a conditional with a True consequent is True.