# What is the predicate of an n-ary propositional function?

I'm self-studying discrete mathematics using the Rosen textbook, and I'm trying to get some predicate logic terms straight. Using definitions from that textbook:

The propositional function P(x) is "x < 3", and has as its subject the variable x, and as its predicate < 3. The predicate does not use the variable.

QUESTION: If Q(x,y,z) is "x = y + z", what is its predicate?

It seem awkward to state the predicate without using the variables. (And not using them does not seem to serve any purpose, since they are only defined within the scope of the propositional function, and so I think they won't interfere with any other propositional functions.)

By analogy with English grammar, I suppose Q(x,y,z) would be considered to have a subject that is a compound subject of the tuple (x,y,z), but that doesn't help with the predicate.

I notice that some binary relational predicates can be stated as the words between the variables, such as "is taller than" or "is parent of", but this will not always work for every n-ary propositional function in general - such as "x = y + z".

Perhaps I'm demanding too much precision in these terms - but precision is the reason I'm studying this, so I'd really like to know. :-) Thanks for any clarification/correction.

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You can think of the predicate as "... is the sum of ... and ...". The fact that you can sometimes get away without putting words in the middle of the predicate is just a quirk of English. –  Zhen Lin Aug 2 '12 at 9:31
I think of predicate as a set. Then we are just thinking of all ordered triples $(x,y,z)$ such that $\dots$. –  André Nicolas Aug 2 '12 at 11:52
Calling "< 3" a predicate is a bit confusing. Less confusing might be calling it the "is less than 3" predicate. –  Dan Christensen Aug 3 '12 at 4:49
The x=y+z might be called the "is an additive triple" predicate. –  Dan Christensen Aug 3 '12 at 4:56
@ZhenLin Thanks, yes, that works to separate the variables and predicate. In effect, the gaps become positional variables, named "..."<sub>1</sub>, "..."<sub>2</sub>, ... , "..."<sub>k</sub>. Why not just use the variables? I note propositional functions are sometimes informally called "predicates", which amounts to this. BTW: I agree it's a quirk - that's why I wanted a general solution. :) –  13ren Aug 3 '12 at 9:06
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## 1 Answer

To be honest, Rosen's book is not reliable when it comes to logical niceties. For example, in the passage you are referring to, he writes

The statement “x is greater than 3” has two parts. The first part, the variable x, is the subject of the statement. The second part—the predicate, “is greater than 3”—refers to a property that the subject of the statement can have.

For a start, “x is greater than 3” isn't a statement in a normal sense of the term (it has a dangling free variable), and -- it we take subject terms in the ordinary sense to be putative referring expressions -- then “x” isn't a subject term either.

As for predicates, how best to think of them (as strings of words, as strings of words with marked gaps, or other alternatives) has been a vexed question ever since Frege. There's a lot initially to be said for the Fregean line that we should treat predicates in the second way -- so for example '$\xi$ killed $\zeta$' and '$\xi$ killed $\xi$' count as distinct predicates (the first murderous, the second suidical!), with '$\xi$' and '$\zeta$' as place markers. But the arguments about the best treatment soon get murky: see this fine article http://www.phil.cam.ac.uk/teaching_staff/oliver/What_is_a_Predicate.pdf

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Thanks! Unfortunately, like papers, textbooks define their own terminology. Rosen defines the term "proposition" as a statement that is either true or false - so what you are calling a "statement", he calls a "proposition". He makes your point that free variables must be bound (assigned a value or quantified) to create a proposition. (The textbook is at the library, so I can't give a page number or quote). I've checked many texts on discrete maths, and I'm unhappy with all of them. Given this, Rosen is OK (though many Amazon reviewers hate it, as they do Johnsonbaugh). –  13ren Aug 3 '12 at 14:08
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