# For arbitrary x: 'Predicate Constant'(x) vs ∃x ( 'Predicate Constant'(x) )

Source: p 348, Understanding Semantics (2 ed, 2013) by Sebastian Löbner

We can now see why employment of an existential quantifier makes a big difference. Compare (31a) with the ‘naked’ (31b):

(31) a. ∃x ( fox( x ) Λ wicked( x ) )
b. fox( x ) Λ wicked( x )

The quantifier-free formula in (27b) is just about a single case, namely about the individual we happened to fix as the interpretation of the variable x in case it is used as a free variable.
In our model, (27b) thus happens to be false because
[1.] [x] is defined as Ken [by this author on p 343], and Ken is neither a fox nor wicked.
[End of 1.]
If (27b) were true, it would be a coincidence.

[2.] The use of a quantifier with a certain variable cancels the value assignment for the bound variable. [End of 2.]
This is why the distinction between free and bound variables is so important: free variables are simply terms denoting a certain individual specified in the model as the value of the variable; bound variables, however, are 'real' variables that stand for all Individuals in the universe.

1. Suppose 1 false: now [x] is not defined by Ken, but [x] means any arbitrary individual. Then:

2. how does (31a) compare with (31b)? Do they still differ?

3. would 2 change? Would no value assignment be cancelled?

• Is your 3 possible in this formalism? If so, why would there be quantifiers, and how would you denote bound variables? Commented May 28, 2016 at 6:34
• @LePressentiment Are you still interested in your question, also your related one? You haven't reacted to any of the answers yet, and I'd like to know if my answers were any helpful. Commented Jun 11, 2016 at 15:03
• @lemontree Sorry for the delay; yes, I will return to you no later than tomorrow.
– user5306
Commented Jun 11, 2016 at 16:14
• There's no hurry :) Just wanted to nudge in case you have simply forgotten the topic. Commented Jun 11, 2016 at 17:55
– user5306
Commented Jun 13, 2016 at 3:52

Making usual assumptions about how the symbols are interpreted, your 31a (with the existential quantifier) is a proposition, which can be true or false, but 31b (without any quantifier for x) is a propositional function, which is a function from individuals to truth values. It does not have a unique truth value.

I don't find the parts you quoted from your authority to be terribly clear, but I think perhaps the "x" of 31b is intended to be a "parameter" in the sense Richmond Thomason uses that term in his textbook Symbolic Logic: An Introduction. A parameter can refer to an individual with the understanding that the choice of which individual it does refer to is arbitrary (or the choice could be subject to assumptions about what predicates hold true of that individual). Thomason uses a different set of symbols to represent parameters, u, v, ..., to keep them apart from variables, x, y, ... and constants, a, b, ....

"x" will never refer to any arbitrary individual, but to exactly that individual that the current variable assignment maps it. So there is no real answer to your question 5, because your supposition just doesn't work.

The reason why in the first sentence, the current variable assignment is "cancelled" is because as soon as we use a quantifier, we abstract away from the current assignment function (let's call this g), and take a look at all the possible other functions to assign to x (x is an individual of the domain) while the assignments for all the other variables remain the same (usually those functions are called the x-alternatives of g).

But this obviously requires the particular assignment functions to uniquely map between variables and elements of the domain. This is the crucial feature of a function: For any given argument, there we always be exactly one (or, if the function is not defined for that argument, no) value for that argument. This definition of the function means that you can't just suppose that x "referred to any arbitrary individual", because this is not what a function does. You need to stick to one mapping and if the individual denoted in that way happened to be a wicked fox, everything is fine; but you can't start ranging over other possible mappings if you don't properly define the meaning of a single assignment.

So, 5 is not really answerable because your pre-assumption just doesn't work.

For question 4:

In case the constants in your 31b happen to denote a wicked fox, and in case (if 31a is fulfilled, 31b necessarily is) the existential quantifier in 31a "finds" a variable assignment which refers to a wicked fox, then indeed the extensions of the two sentences are the same, namely true.
The same holds if there is just no wicked fox in the actual world, in that case both sentences will come out as false.
However, it might happen that in fact there is an individual which is both a fox and a wicked one, but that x in the second example just doesn't refer to that fox, but, as in the example, e.g. to Ken. In that case, the two sentences do differ: 31a will still be true if there is just some individual which is both a fox and wicked, but 31b will fail if we choose a variable assignment which assigns x an individual which is not a fox, but Ken - or maybe a fox, but not a wicked one, or vice versa; that may as well happen (due to the truth conditions of the conjunction).

In any case, even if the extensions of the whole sentences are equivalent, so are not the extensions of the individuals it involves: 31a, employing a quantifier, will always involve a set of individuals - no matter if that set is empty (because there is no such thing in our domain that is a fox and wicked), a singleton set (because there is exactly one wicked fox) or in the most extreme case even containing every individual there is.
31b however makes use of a free variable, which behaves like a constant, and always refers to one particular individual.
I.e., even there happens to be only one fox in our actual world that makes the sentence true, the logical types within the formulas still differ.