Quantified Modal Logic - How does an existence predicate help with a fixed-domain model?

When constructing models for a first order modal logic, we have to decide whether we want to have a fixed domain (the possibilist semantics) or a world-relative domain (the actualist semantics).

In the fixed-domain model, the quantifiers range over all possible objects. Some will introduce an existence predicate, in order to restrict statements to the "actual" world.

The SEP page on modal logic summarizes the situation like this:

Furthermore, those quantifier expressions of natural language whose domain is world (or time) dependent can be expressed using the fixed-domain quantifier ∃x and a predicate letter E with the reading ‘actually exists’. For example, instead of translating ‘Some Man exists who Signed the Declaration of Independence’ by

∃x(Mx & Sx),

the defender of fixed domains may write:

∃x(Ex & Mx & Sx),

thus ensuring the translation is counted false at the present time.

I'm having trouble grasping how "∃x(Ex & Mx & Sx)" fixes an interpretation to the actual world any more than "∃x(Mx & Sx)" does.

As I understand it, these expressions are always evaluated at a world. So "Ex" will only be true in worlds where Ex is satisfied. (For instance, assume that G is an interpretation function, and assume that "Ex" is only satisfied in worlds where G(x) \in G(E).)

But wouldn't the same be true for "Mx"? Why should "Ex" be satisfied in one world any more than "Mx"?

What am I not grasping here?

My reading of the Modal Logic entry of SEP is that it does not claim that "$\exists x (Ex \wedge Mx \wedge Sx)$" fixes an interpretation to the actual world.

Introduction of the predicate letter $E$ merely gives an adopter of constant-domain quantification the ability to write a sentence asserting the existence of a man who signed the Declaration of Independence that may be false in one world and true in another, accessible world.

Interpreting $Ex$ as "$x$ actually exists in this world" makes the sentence false in this world and true in the world of 1777. (Provided the obvious interpretations are given to $M$ and $S$ too.)

In 1775, however, no man had signed the Declaration of Independence. The sentence "$\exists x (Ex \wedge Mx \wedge Sx)$" is false in that world too, though the interpretation is not fixed to the actual world.

• Is "∃x(Mx & Sx)" true in the 1775 world? I would have thought that it isn't, because Mx and Sx won't turn out true in that world (assuming no object from the domain shows up in the interpretation of M and S in the 1775 world). – user1439929 Feb 12 '18 at 12:59
• In constant-domain quantification, free variables take as values objects from the domain of the model, not necessarily from the domain of the "current world." Hence $\exists x(Sx)$ is true in the 1775 world if there exist some element $x$ of the domain of the model (say, in the 1777 world) for which $Sx$ is true. – Fabio Somenzi Feb 12 '18 at 15:36
• I must be confused. I’m having a hard time understanding how ∃x(Sx) could be true in the 1775 world, when Sx is not satisfied in the 1775 world. – user1439929 Feb 14 '18 at 1:44
• Let $w1775$ and $w1777$ be two worlds. Let the sole domain $D$ contain two objects $\{ m, n \}$. Let $G$ be an interpretation of the predicate $S$ such that $G(S, w1775) = \{\}$ (which means nothing satisfies $S$ in $w1775$), and $G(S, w1777) = \{ m \}$ (which means $m$ satisfies $S$ in $w1777$). Under this interpretation, the truth conditions for predicates are relativized to worlds, aren’t they? So $\exists x(Sx)$ should be true only in $w1777$, and not in $w1775$. Or am I mis understanding something? – user1439929 Feb 14 '18 at 1:47
• Yes, when you define a model you choose the interpretation of each relation symbol in each world of the model. In the constant-domain approach, the interpretation of a relation symbol is a relation of the proper arity on the domain of the model. That's exactly what happens to the $E$ unary relation symbol. Its introduction frees one from having to say that Benjamin Franklin is not a man or is not a signer of the US Declaration of Independence, given that Franklin is an element of the model's domain. (To be continued.) – Fabio Somenzi Feb 14 '18 at 2:51