Element contained within a predicate?? Tattoo error...

Question

Hey so my friend got a tattoo of logical symbols to translate some quotation, and this friend ended up having a statement of this form contained in the proposition: $\forall y[P(y) \rightarrow (y \notin Q(x))]$ I saw this as problematic, as P an Q being unary predicates, they are a relation operation and not a set of elements, so $y\notin Q(x)$ makes no sense. Let me know if this is valid or not, it's no my tattoo but my friend would very much like to know. Thanks

Answer 1

Since $x$ is not bound, this expression is a function of $x$. It makes sense, but it is not a statement. A statement would be $\forall x$ (this expression), or $\exists x$ (this expression), or $\forall x\forall z,~ (z\to x)\to$ (this expression).

Answer 2

Here, I believe that what is meant is for $$Q(x) = \{x\in \text{domain}\mid Q(x) \,\text{is true} \}$$

So I suspect that the intent in declaring any element $y$ to not to be an element of $Q(x)$ was to mean, in effect, $\lnot Q(y)$.

So taking $y$ to be in the relevant domain, I suspect what was meant was

$$\forall y(P(y) \rightarrow \lnot Q(y))$$

But as noted in Andreas' comment below, there is a problem with this interpretation, in that we really do not have any bound on $x$ in its appearance in the tattoo, and its appearance on the left of the above attempt to make sense of what is meant by the "set" $Q(x)$.

Answer 3

In order for "$y\in Q(x)$" to make sense, $Q(x)$ will have to be a set, and therefore $Q$ will have to be a set-valued function, not a predicate. Interpreting $Q$ this way (and interpreting $P$ as a unary predicate), the formula is OK; its truth value will depend not only on these interpretations of $P$ and $Q$ but also on the value of the free variable $x$. Whether it corresponds to the quotation your friend had in mind is another question (which I fear is likely to have a negative answer).