|
Modal Predicate Logic |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
Introduction |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
The modal logics looked at so far have been extensions of the standard propositional logic. Now we will convert a modal propositional logic to a modal predicate logic in just the same way that we converted the standard propositional to standard predicate. We will also modify the semantics - by which I basically mean here the truth trees - in the same way that truth tree rules for PL were modified for QL. We will restrict ourselves to a quantified adaptation of S5. We'll also be concerned to add the special predicate '=' to the language, which we'll naturally think of as the predicate representing 'identity'. There are several ways to do this.
There are many philosophical problems associated with quantified modal logics. There are questions concerning the nature of instantiations across worlds, there are questions concerning the identity of items that are referenced in the scope of quantifiers, there are questions about the ontological assumptions that may be built into the semantics of the logics. We'll look at some of these problems later
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
S5QT: A Modification of S5 |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
Tree Rules
Add the following rules to the rules for S5 given previously (i.e. PL + MN + {<>S5, []S5}) to get S5QT:
QN ~($x)X w/ ... ("x)~X w
~("x)X w/ ... ($x)~X w/
EI ($x)X w/a ... X' w where a is new to this branch, and X' is the result of replacing every free occurrence of x in X by a
UI ("x)X w\a ... X' w where a is any constant, and X' is the result of replacing every free occurrence of x in X by a
Show that the Barcan Formula is S5QT-valid: ("x)[]Fx É []("x)Fx
It is also worth noting that the following formulae are also S5QT-valid []("x)Fx É ("x)[]Fx ($x)<>Fx É <>($x)Fx <>($x)Fx É ($x)<>Fx
In fact all the last four are actually KQT-valid, which means that in any of our normal logics [] and (x) are commutative, as are <> and ($x).
On the other hand, try this formula: <>(x)Fx É (x)<>Fx
Counterexamples
Try this formula: (x)<>Fx É <>(x)Fx
1 ~((x)<>Fx É <>(x)Fx) n NTF 2 (x)<>Fx \a, b n 1 3 ~<>(x)Fx n 1 4 []~(x)Fx n 3, MN 5 <>Fa \k n 2, UI 6 Fa k 5, <>S5 7 ~(x)Fx k 4, []S5 8 ($x)~Fx k 7, QN 9 ~Fb k 8, EI 10 <>Fb n 2, UI 11 Fb m 10, <>S5
It's invalid - which you will have expected if you thought about what it was saying.
Create the counterexample from this tree
We can show this is a counterexample by expanding the formula for domain {a, b} and assigning values in n
Find a counterexample for this formula: []($x)Fx É ($x)[]Fx
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
Identity |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
Tree Rules
We add identity rules to get S5QT=:
CSI x = y w X w ... X' w
where X' is the result of replacing any number of occurrences of x or y in X by y or x respectively [CSI is for contingent substitutivity of idenicals]
Closure a ≠ a w ... X
Note that as defined CSI only marks objects as being the same in a world. In such a case the following would not necessarily lead to a closure:
a = b n a ≠ b k
Instead of CSI we could add a rule that allows identification across worlds to get S5QT[]=, thus:
NSI x = y v X w ... X' w
where v is any world index, and X' is the result of replacing any number of occurrences of x or y in X by y or x respectively
Test this formula for validity in S5QT= and S5QT[]=: (a=b) É [](a=b)
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
Existential Import |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
One of the fundamental problems that arise in quantified modal logic is the problem of existential import. Does the very nature of the logic commit you to claims about what things must exist in the world? and how?
Quantifiers
We can interpret quantifiers as with or without existential import. And we can read them to ourselves differently to remind us of the interpretation we're adopting.
a. Without existential import: ($x)Fx is a particular quantifier, read as 'at least one thing' is F - and that thing may or may not exist. (x)Fx reads as 'everything' is F
b. With existential import: ($x)Fx is an existential quantifier, read as 'there exists at least one thing' that is F. (x)Fx reads as 'everything there is' is F (or, given the QN equivalences, you could read it as 'non-Fs do not exist'.)
Problems:
a. With Ex. Imp. we can't translate 'Some things do not exist' meaning things like Pegasus, not that 'nothing exists' You can't say ($x)~Ex : there exists at least one thing that doesn't exist. That's a contradiction.
b. With Ex. Imp. we translate ($x)(x = a) as 'There exists at least one thing identical to a'. So: Let a = Mister Pickwick, then we translate ~($x)(x = a) as 'Mr. Pickwick does not exist.' but ($x)(x = a) is S5QT=-valid, so the negation would be a S5-contradiction. But the English sentence doesn't look contradictory
Constants
Instead of supposing that only the quantifiers have existential import, suppose that the constants have it too. Consider Fa É ($x)Fx. It is QT-valid. Read it as 'if a is F, then there exists at least one thing that is F' Here 'a' must refer to an existing thing. This is typical of standard logics: constants have ex. imp. if quantifiers do. Therefore constants can't be used to refer to non-existents in standard logic with ex. imp.
Consider now the Barcan Formulae in S5QT: <>($x)Fx É ($x)<>Fx read as 'if it is possible that there exists an F, then there exists a possible F' This is hard to interpret - especially the consequent of the conditional. A possible F? At least one of them exists? Try interpreting <> as a possible world indicator, then read as 'if in some possible world there exists an F, then there exists something that in some possible world is an F' But this also raises questions: 1. Does it say there is a thing in this world which is an F in some world? 2. Does it say in some world there is a thing in this world which is an F in some world? 3. Do all entities labeled by a constant in each world exist in that world? 4. Is the same set of existents in each world but with different properties
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
Domains of Quantification |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
The other fundamental problem that arises in quantified modal logic is the problem of domains of quantification. The domain the the set of things that are referred to by the quantifiers. If the quantifiers have existential import then the domain is a set of existing things.
In quantified modal logic we need to know whether the domain is the same in every world, or whether it is different in different worlds. There are three general situations.
AE: every world has a distinct non-overlapping domain. BE: every world has the same domain. CE: every world has its own domain which may or may not overlap
For CE there are two special situations:
CEA: if wAv then the domain in w is a subset of the domain in v. 'A' is for 'ascending'. CED: if wAv then the domain in v is a subset of the domain in w. 'A' is for 'descending'.
Note that for S5QT= CEA and CED both collapse to BE. Assuming BE we can easily understand <>($x)Fx being equivalent to ($x)<>Fx. Still, with existential import we can't have <>($x)(x = a), because it is still contradictory.
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
Modality and Existence |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
We want a logic to match CE. We use modifications of the UI rules to create variations on modal predicate logics. Construct RMQT= by introducing RUI as follows:
RUI ("x)X w\a ... X' w where a is any constant that occurs in a formula in w on this branch, or a is new to this branch, and X' is the result of replacing every free occurrence of x in X by a
Show that the Barcan Formula is not RMQT=-valid: ("x)[]Fx É []("x)Fx
In fact, it's even open in S5.
From the truth tree above - and considering it as a tree in S5 now - we can construct a counter-example (If it's a c-ex in S5 then it's a c-ex in all the other normal logics we've looked at.)
For ("x)[]Fx (n) eliminate the quantifier as we do for normal predicate logic, to get []Fb (n). So ((x)[]Fx É [](x)Fx) (n) = 1 É 0 = 0
In this system all the quantifiers are relative to worlds. So
("x)[]Fx (n) is to be read as 'Everything that exists in n is F in all worlds to which n has access' and []("x)Fx (n) is to be read as 'In all worlds accessible from n, everything that exists is F,'
Consider the two worlds n and k above. The domain of n is just {b}, and the domain of k is {a, b} We can see that the antecedent of BF is true and the consequent is false.
On the other hand, if in RMQT= we test the converse of the BF [](x)Fx É (x)[]Fx we will find that the tree closes in K, which is reasonable given the reading of the converse.
What we've assumed so far is that if a constant 'a' occurs in a formula in a world n, then 'a' exists in n. That's what RUI plays on. But a formula like <>~Fa (n) is interpretable in the possible worlds style as saying that there is no world accessible from n in which a isn't F, and this interpretation doesn't make us want to assume that a does exist in n. We can remove this assumption by making RUI Tighter, thus:
TRUI ("x)X w\a ... X' w where a is any constant that occurs in a formula in w on this branch, but not in the scope of a <>, or a is new to this branch, and X' is the result of replacing every free occurrence of x in X by a
However, if in TRMQT= we again test the converse of the BF [](x)Fx É (x)[]Fx we will find that the tree still closes in K.
We can stop both that and the BF itself from being valid by using the Harshly Restrictive UI. thus:
HRUI ("x)X w\a ... X' w where a is any constant that occurs in a formula in w on this branch, but has not been used for EI in any world but w, or a is new to this branch, and X' is the result of replacing every free occurrence of x in X by a
If in HRMQT= we again test the converse of the BF [](x)Fx É (x)[]Fx we will find that the tree does not close, even in S5.
In the counterexample that we construct from this tree, a doesn't exist in k, therefore ~Fa (k) is set to 1
Because each world is accessible from each other world (assuming S5 rules) and a Ï k and b Ï n, we have that []Fa and []Fb are both 0.
Given this interpretation of the quantifiers the antecedent and the consequent of the converse BF are to be read thus:
("x)[]Fx (n) is to be read as 'Everything that exists in n is F in all worlds to which n has access' and []("x)Fx (n) is to be read as 'In all worlds accessible from n, everything that exists in n is F, if it exists there at all.'
Notice that it is the interpretation of the quantifiers that allows us to construct different counter-examples, or even to be able to find counter-examples.
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Suppressing Existential Import | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
In the previous section we assumed existential import for quantifiers and constants. This made us interested in details of the Domains for the different worlds. But we don't need to think of a Domain in a world as all the existing things in that world. A Domain could be all the things - some existing, some not. That would allow us to suppose that all the domains are identical; but that some elements exist in n and some exist in k.
We can read the BF more simply now as:
BF: If everything in all worlds is F, then in all worlds everything is F Converse BF: If in all worlds everything is F, then everything in all worlds is F.
These are going to be valid using straightforward UI if everything means everything not just every existing thing.
How would we translate the interpretation of the BF formula that we came up with for RMQT= if we don't have existential import? Remember it was
BF: If everything that exists in n is F in all worlds to which n has access, then in all worlds accessible from n, everything that exists is F.
We will need to be able to distinguish those things that the quantifiers (one of which is read as 'there exists') quantify over, from those things which actually exist. We use a predicate 'Ex' to say 'x exists'.
Try this formula: (x)<>(Ex É []Fx) É (x)[](Ex É Fx)
The tree that we get (assume S5 at the end - and all words accessible to all worlds) gives us a counterexample:
(Eliminate quantifiers for a to prove it is a c-ex.)
Identity and Existence
Remember that the discussion of identity mentioned that the wff ~($x)(x = a) was a contradiction
1 ~($x)(x = a) n NTF 2 (x)~(x = a) n 1, QN 3 ~(a = a) n 2, UI, a X
But earlier, we allowed the assumption that if a doesn't exist in k, then ~Fa (k) may be set to 1. Now, in this case, if a doesn't exist in n, then perhaps ~(a = a) shouldn't be contradictory. What we need to do is to make the closure rule apply only if non-self-identity is asserted of constants that are known to exist in the world. A rule to make that clear is:
RClosure ($x)(x = a) w ... a ≠ a w X
But since ($x)(x = a) is valid (see tree above) the rule RClosure will be apply everywhere anyway. In order for RClosure to make a difference we have to remove existential import from the constants. This gives us Free Logic.
Free Logic
In FL the quantifiers have Ex. Imp. but not the constants, so Fa É ($x)Fx is not valid.
Here are quantifier rules to formalise this interpretation:
FEI ($x)X w/a ... X' w ($x)(x = a) w where a is new to this branch, and X' is the result of replacing every free occurrence of x in X by a
FUI ("x)X w\a ($y)(y = a) w ... X' w where X' is the result of replacing every free occurrence of x in X by a
Try this formula: ($x)(x = a)
1 ~($x)(x = a) n NTF 2 (x)~(x = a) n 1, QN
And that's as far as we can go.
What's happening with FEI and FUI is that the domain of quantification is being restricted. Not all the items in the world are available for the quantifiers. In particular the FUI applies only to existing things in the world. But that's OK because we'll say that the universal quantification of Fx is true in w iff Fa is true for all a that exist in w. And, of course, if the domain of quantification is empty then all universal quantifications are vacuously true.
Now try the BF and converse again. They both turn out to be open and we get the following counterexamples in S5:
For BF
Note that nothing has been declared to exist in n, (i.e. Q(n) = Æ) so all universal quantifications in n are vacuously true.
For the converse of BF
Note that nothing has been declared to exist in k, (i.e. Q(k) = Æ) so all universal quantifications in k are vacuously true.
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||