I'm having trouble creating models for predicate logic statements. I am going to give an example, I hope you guys can help me out.
$$\forall x(P(x) \text{^} Q(x,a) \to Q(x,b) )$$
Let $M_1$ be the following model:
- dom($M_1$) = N
- $P ^M (x) =x$ is positive
- $Q^M(x, y) = x \ge y$
- $a^M = 5$
- $b^M = 0$
To my understanding this means that for all v $ \in N $, if $v$ is positive and $v \ge 5$ then $v \ge 0$.
$M_1 \models \forall x(P(x) \text{^} Q(x,a) \to Q(x,b) )$
Is this correct?
Now say I'm trying to make this statement false.
Say I were to switch constants a and b, such that $a^M = 0$ $b^M = 5$. Would that mean $M_1 \nvDash \forall x(P(x) \text{^} Q(x,a) \to Q(x,b) )$?
Any help would be greatly appreciated. I'm really stuck with this. Thank you in advance.
Just as a further suggestion for when you encounter this sort of problem: the easiest thing to do is to choose models with finite domains. I'd even say try and put as few elements as possible in them. Remember that the domains and the denotations of predicates are just sets, and as such you can always define them by extension (there is no need to look for a property which members of the set satisfy -that is, no need to define them by comprehension).
For instance, in the excercise you just posted, I would have just done this:
Since the antecedent of the conditional is always false, the sentence will always be true.
Another option would have been $P^M$(x) = dom(M1), and $Q^M$(x,y) = dom(M1)$^2$ (just have everything satisfy property P, and everything relate to everything through Q)
If you need to make the sentence false, then have some element be a P, and relate through Q to $a$ but not to $b$. For instance $P^M$(x) = {1}, and $Q^M$(x,y) = {<1,1>}
The method will not always work (there are sentences and sets of sentences which can only be satisfied by infinite models), but in many cases it will, and I think is a good 'first approach' possible.