In other words, does the universe of discourse limit the interpretation of the predicate? So for example, say the universe of discourse is $\mathbb{Z}^+$ (positive integers). Let $P(x)$ be "$\sin(x) > 0$". If $x = 1$ then $\sin(x)$ is no longer in $\mathbb{Z}^+$. Does that mean $P(x)$ is false for those values of $x$ even though the inequality is true if you allow reals in the calculation? Or does the predicate exist independent of the universe of discourse and that universe applies only to the variables?

NOTE: if the universe of discourse was not stated explicitly or if we had a statement like $(\exists x\in\mathbb{Z}^+ \;|\; \sin(x) > 0)$ then I would not be concerned about the predicate "leaving" $\mathbb{Z}^+$.

PS. I've looked through all the "Similar Questions" that come up in the sidebar, I've searched for a bunch of variations on "predicate universe of discourse", and of course lots of Google searching, but can't seem to find a conclusive answer. I started getting in to ZFC axioms to try to sort this out, but that wasn't going anywhere either.

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    $\begingroup$ "does the universe of discourse limit the interpretation of the predicate?" Yes, of course: the interpretation of a predicate $P(x)$ must be a subset of the domain. $\endgroup$ Oct 10 '18 at 18:35
  • $\begingroup$ Simple example: the sentence $\forall x (x \ge 0)$ is true in $\mathbb N$ and false in $\mathbb Z$. $\endgroup$ Oct 10 '18 at 18:38
  • $\begingroup$ Perhaps clarifying: can a predicate contain symbols or elements which are outside the universe of discourse. E.g. if the universe of discourse is ℤ+, does the predicate x > 0 even make sense (zero not being in ℤ+)? $\endgroup$ Oct 10 '18 at 18:46
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    $\begingroup$ In the formula $x > 0$ $0$ is a constant symbols that must denote an element of the domain. If not, the formula has no meaning. $\endgroup$ Oct 10 '18 at 19:34
  • $\begingroup$ plato.stanford.edu/entries/logic-free $\endgroup$
    – DanielV
    Oct 11 '18 at 7:17

The question is: how are you defining/axiomatizing $\sin$? It is unlikely you can fully characterize $\sin$ in a theory where $\mathbb Z^+$ is a valid universe.

Ignoring that, let's say you just have a function symbol that you are calling $\sin$ and which you want to interpret as the function $\sin$. Then, if the universe was $\mathbb Z^+$, it would need to be a function $\mathbb Z^+ \to \mathbb Z^+$, but the only integer $\sin$ takes to an integer is $0$. Therefore this would simply not be a valid interpretation. (I'm also assuming $\mathbb Z^+$ contains $0$. If not, then you just end up with a totally undefined "function" and an empty relation below.)

You could instead use a predicate formulation such as $\sin(x,y)$ to mean intuitively mean $\sin(x)=y$, then you would not have an issue. This binary predicate symbol would be interpreted as a relation containing only $(0,0)$. With this predicate symbol, you could reformulate your formula as $\exists x.\forall y.\sin(x,y)\land y>0$. This would be false in the interpretation I suggested since $\sin(x,y)$ is only true for $x=0$ and $y=0$.

  • $\begingroup$ For clarification, I was implicitly defining it in the "normal" sense on the reals. There are other ways of describing the problem... for example if the universe is ℤ+ and the predicate is (x-5)+5 then we "leave" ℤ+ in calculating inside the first parentheses (because subtraction is not closed on ℤ+). So in this case, the focus of my question is this sense of "leaving" the universe of discourse, and if that is "allowed" or if doing so makes the predicate undefined or invalid in some sense. $\endgroup$ Oct 10 '18 at 18:57
  • $\begingroup$ The issues are the same. You'd still need to define/axiomatize subtraction. Using the standard subtraction makes your interpretation of the function symbols not be functions on the universe. This makes the interpretation as a whole not an interpretation of the particular theory. You can deal with partially defined operations by using a relational approach such as I described. Interpreting a function symbol as a partially defined function is simply not allowed (for standard semantics), so you never even get to the point where you can ask whether a formula is true or not. $\endgroup$ Oct 10 '18 at 19:03

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