Multi-Place Predicates and Multiple Quantifiers

Transcription

1 Multi-Place Predicates and Multiple Quantifiers So far we have looked only at monadic predicates. That is, predicates that apply to only one thing. For instance, Fa might mean Alice is friendly. But, what about predicates that involve TWO individuals otherwise known as relations for instance, Alice loves Burt? We will now introduce a way to symbolize relations: Sentence Symbolization Alice loves Burt. Lab Burt loves Alice. Lba Alice does not love Burt. Lab If Alice loves Burt, then Burt loves Alice. Lab Lba Burt loves himself. Lbb Alice loves something. (ꓱx)Lax Something loves Alice. (ꓱx)Lxa Alice loves everything. (ꓯx)Lax Alice loves someone. (ꓱx)Lax * Everyone loves Alice. (ꓯx)Lxa ** If anyone loves Alice, then Burt loves Alice. (ꓱx)Lxa Lba *** * Technically, this says Alice loves something. When we say Alice loves someone we do not mean just any old thing, but some PERSON. Hence, we really ought to interpret the symbolizations of the starred statements as follows: * (ꓱx)(Px Lax) There exists an x such that x is a person and Alice loves x. ** (ꓯx)(Px Lxa) For all x, if x is a person, then x loves Alice. *** (ꓱx)(Px Lxa) Lba If there exists an x such that x is a person and x loves Alice, then Burt loves Alice. But, for simplicity, for propositions like this, we will assume that, e.g., (ꓯx) refers to a restricted domain, and means something like for all PEOPLE rather than for all THINGS. We can also express 3-place relations (and 4-place, 5-place, and so on). Alice passed the salt from Burt to Casey. Burt stood between Alice and Casey. Alice believes that Burt drove Casey to Daphne s place. Pabc Sbac Babcd 1

2 Here s a tricky one, because it involves two quantifiers (and two variables): (1) Someone loves everyone. (ꓱx)(ꓯy)Lxy We need two DIFFERENT variables here. What the symbolization literally says is: (1) There exists an x such that, for all y, x loves y. Note: The statement (ꓱx)(ꓯx)Lxx is not a wff. Note that, according the criteria for binding, the existential quantifier doesn t bind anything! Here are a few more: (2) Someone is loved by everyone. (ꓱx)(ꓯy)Lyx There exists an x such that, for all y, y loves x. (3) Everyone loves someone. (ꓯx)(ꓱy)Lxy For all x, there exists a y such that x loves y. (4) Everyone is loved by someone. (ꓯx)(ꓱy)Lyx For all x, there exists a y such that y loves x. Here are two complicated sentences involving THREE quantifiers: (5) Everyone loves at least one person (ꓯx)(ꓱy)[Lxy (ꓯz)Lyz] who loves everyone. Translation: For all x, there exists some y such that x loves y, and, for all z, y loves z. (6) No one loves anyone who loves no one. (ꓱx)(ꓱy)[Lxy (ꓯz) Lyz] Translation: It is not the case that there exists an x and there exists a y such that x loves y, and, for all z, y does not love z. (7) Every unicorn is friends with a dragon. (ꓯx)[Ux (ꓱy)(Dy Fxy)] Translation: For all x, if x is a unicorn, then there exists a y such that y is a dragon and x is friends with y. 2

3 The Finite Universe (Counterexample) Method for Multiples Previously, we learned the finite-universe method of demonstrating invalidity for singlepredicate wff s. We can easily extend that method to wff s with multi-place predicates and multiple quantifiers. Consider, for instance: (ꓯx)(ꓱy)Lxy. The main operator here is the universal quantifier, ꓯ. Remember that ꓯ s are disguised conjunctions, so the main operator of our translation will be. But, we also learned that ꓱ s are just disguised disjunctions, so our translation will be a conjunction of disjunctions: (ꓯx)(ꓱy)Lxy for a domain with ONE object {a} in it: Laa (for one-object domains, just replace all variables with that object s letter; Δxy Δaa) (ꓯx)(ꓱy)Lxy for a domain with TWO objects {a, b} in it: (Laa Lab) (Lba Lbb) Say that (ꓯx)(ꓱy)Lxy means Everything loves something. In a two-object universe there is only a and b (say, Anne and Brett). Everything loves something means that Anne loves something AND Brett loves something. In a two-object universe, the something that they love must either be his/herself or the other person. That means Anne either loves herself or Brett; and Brett either loves Anne or himself. Thus, the interpretation above, which is a conjunction of disjunctions: (Anne either loves herself, or she loves Brett) AND (Brett either loves Anne, or himself) (ꓯx)(ꓱy)Lxy for a domain with THREE objects {a, b, c} in it: (Laa Lab Lac) (Lba Lbb Lbc) (Lca Lcb Lcc) * * technically, there should be lots of brackets here Imagine a universe with only three objects in it: Anne, Brett, and a cookie. If Everything loves something is true, then: (Anne loves either herself, Brett, or the cookie) AND (Brett loves either Anne, himself, or the cookie) AND (the cookie loves either Anne or Brett or itself) Sure, it would have to be a sentient cookie for the third conjunct to come out true, but you get the point. 3

4 Example: Let s do an example. Use the finite-universe method to demonstrate that the following inference is invalid: (1) Everyone loves someone. (2) Therefore, someone loves everyone. First, let s translate these into symbolic form. (Note: For simplicity, we will ignore the fact that it says everyone instead of everything ; let us stipulate that the context of this inference is one where the domain under consideration includes only people in it.) (1) (ꓯx)(ꓱy)Lxy (2) (ꓱx)(ꓯy)Lxy In a two-object universe {a, b}, we would translate these as: (1) (Laa Lab) (Lba Lbb) (2) (Laa Lab) (Lba Lbb) Let s try the following interpretation: Domain: {Anne, Brett} Lxy = x loves y a = Anne b = Brett True statements: Lab ; Laa ; Lba ; Lbb [Basically, this is a universe where Anne and Brett love each other, but not themselves.] Does the premise come out true while the conclusion comes out false? Yes! (Laa Lab) (Lba Lbb) F T T T T F F (Laa Lab) (Lba Lbb) F F T F T F F In this universe, consisting of only two people, everyone is loved by someone (namely, Anne is loved by Brett and Brett is loved by Anne), but no one loves everyone (since neither Anne nor Brett love themselves). And, if the inference above is invalid in ONE conceivable universe, then it is invalid in ALL possible universes (including this one). So, the inference is invalid. Hooray! You ve learned how to use to finite-universe method to demonstrate invalidity for multi-place predicates and multiple quantifiers. 4

5 Derivation Rules for Multiples Note the following additions to the derivation rules: 1. ꓱI for Multi-Place Predicates: Here are some new ways to use ꓱI: 1 (1) Laa A 1 (2) (ꓱx)Lxa 1, ꓱI 1 (3) (ꓱx)Lax 1, ꓱI 1 (4) (ꓱx)Lxx 1, ꓱI Lines (2) (4) are all permissible uses of ꓱI. Line (1) might say Alice loves herself. From this it should be obvious that these statements follow from (1): 1 (1) Laa A Alice loves herself. 1 (2) (ꓱx)Lxa 1, ꓱI Something loves Alice. 1 (3) (ꓱx)Lax 1, ꓱI Alice loves something. 1 (4) (ꓱx)Lxx 1, ꓱI Something loves itself. Note that we might also decide to replace each a one at a time. So, the following inference to line (3) is also permissible: 1 (1) Laa A Alice loves herself. 1 (2) (ꓱy)Lay 1, ꓱI Alice loves something. 1 (3) (ꓱx)(ꓱy)Lxy 2, ꓱI Something loves something. But, sometimes, you ll be introducing an existential quantifier to a wff with free variables in it. In such instances, if Δ contains TWO or more DIFFERENT variables, then you CANNOT replace a variable with one that Δ already contains. For instance, consider: 1 (1) Lxy A 1 (2) (ꓱy)Lyy 1, ꓱI This is NOT permissible because here, you are replacing the x in the original statement with a y, even though the original statement already contained an unbound y. However, the following moves ARE permissible, because we are replacing x with either itself or some variable that the original statement does not already contain: 1 (1) Lxy A 1 (2) (ꓱx)Lxy 1, ꓱI * 1 (3) (ꓱz)Lzy 1, ꓱI * * Restriction: If the formula being bound to ꓱ contains multiple free variables, the variable chosen to be bound to ꓱ cannot be the same as any of the other remaining free variables. 5

6 2. ꓯE for Multiple Operators: When there are multiple operators, you may only eliminate an operator if it is the MAIN (i.e., leftmost) operator. So, for instance, the following move is NOT permissible: 1 (1) (ꓯx)(ꓯy)Lxy A 1 (2) (ꓯx)Lxy 1, ꓯE 3. ꓱE When There Are Multiple Quantified Assumptions: Keep in mind that the following restriction still holds: Last week, we said that when using ꓱE [replacing (ꓱx)Δx with Δx for the purposes of our assumption to be used for ꓱE], we could replace x with any variable we wanted, so long as x did not occur freely in any prior assumption. So, the following is NOT permissible: 1 (1) (ꓱx)Sx A 2 (2) (ꓱx)Cx A 3 (3) Sx Ass. (ꓱE) 4 (4) Cx Ass. (ꓱE) XXX Here, TWO different assumptions are being made for the purposes of ꓱE one in line (3) to eliminate the ꓱ from line (1), and another in line (4) to eliminate the ꓱ in line (2). But, assumptions (3) and (4) should NOT make use of the same variable. Why not? After all, the bound variable in line (1) is an x AND the bound variable in line (2) is ALSO an x. Well, to illustrate why line (4) should have used a different variable, consider what would be derivable if we continue the mistaken proof above: 1 (1) (ꓱx)Sx A 2 (2) (ꓱx)Cx A 3 (3) Sx Ass. (ꓱE) 4 (4) Cx Ass. (ꓱE) XXX 3, 4 (5) Sx Cx 3, 4, I 3, 4 (6) (ꓱx)(Sx Cx) 5, ꓱI 3 (7) Cx (ꓱx)(Sx Cx) 4, 6, I 2, 3 (8) (ꓱx)(Sx Cx) 2, 7, ꓱE 2 (9) Sx (ꓱx)(Sx Cx) 3, 8, I 1, 2 (10) (ꓱx)(Sx Cx) 1, 9, ꓱE Imagine that line (1) says that Something is a snake and that line (2) says that Something is a cat. If the derivation above were permissible, we d have just demonstrated that we can derive Something is both a snake and a cat from these two premises clearly this does not follow. 6

7 Why did this mistake happen? Well, line (1) says that SOMETHING is a snake. Meanwhile, line (2) says that SOMETHING is a cat just because both premises make use of x as their variables, this does NOT mean that lines (1) and (2) are referring to ONE AND THE SAME SOMETHING. Imagine that I told you that Something is a snake and Something is cat. It would be absurd for you to conclude that, Because you used exactly the same word in each claim namely something I can conclude that you were referring to one and the same thing each time. Similarly, just because both premises make use of an x, this does not indicate that the x refers to one and the same thing in each premise. In short, the rule here is that, when choosing a variable to use in our assumption for the purpose of ꓱE, that variable cannot occur freely in ANY other prior assumption. So, because x occurs freely in the assumption on line (3), it is not permissible to make use of x in our assumption for ꓱE on line (4). Note #1: Note that this only applies to the variable that is bound to the ꓱ. So, the following move IS permissible: 1 (1) (ꓱx)Lxy A 1 (2) Lxy Ass. (ꓱE) * Notice that (2) DOES contain a free variable that DOES occur freely in some prior assumption; namely, y occurs freely in both (1) and (2). But, the assumption in line (2) is permissible because the restriction only applies to the variable that is bound to the ꓱ that is being eliminated. So, since x is the variable that is bound to the ꓱ in line (1), it is x that cannot occur freely in any prior assumption when we assume Lxy on line (2). Note #2: Note that a similar thing can be said of both ꓯE and ꓯI. The following uses of ꓯE and ꓯI are permissible: 1 (1) (ꓯx)Lxy A 1 (2) Lxy 1, ꓯE 1 (3) (ꓯx)Lxy 2, ꓯI Our universal elimination on line (2) DOES contain a variable that occurs freely in the assumption that it rests on (namely, y ). Similarly, so does our universal introduction on line (3). However, both are permissible because the restriction on ꓯE only applies to the variable that is being UNBOUND FROM the ꓯ, and the restriction on ꓯI E only applies to the variable that is being BOUND TO the ꓯ. E.g., since we are universalizing x in line (3) that is, we are binding only the x to the universal quantifier it is only the x that cannot occur freely in the assumption that it rests on i.e., line (1). 7 * In short, the restriction only applies to the variable that BECOMES free in the ꓱE assumption.

8 4. ꓯE for Multi-Place Predicates: Formerly, we said that when using ꓯE [e.g., replacing (ꓯx)Δx with Δx], we could replace x with any variable unless that variable occurred freely in the assumption that (ꓯx)Δx rests on. Now we will add another restriction: IF the Δ being stripped of its ꓯ falls under the scope of MORE THAN ONE operator, you may NOT pick a variable that is bound by one of those other operators (but this applies only if your chosen variable would become bound by one of those other quantifiers). Confused? Let s look at an example. The following move is NOT permissible: 1 (1) (ꓯx)(ꓱy)Lxy A 1 (2) (ꓱy)Lyy 1, ꓯE You cannot replace x with y here because y is already in use i.e., it is bound by ꓱ in the original statement that we are eliminating the ꓯ from. When using ꓯE, the variable that was bound to ꓯ should become a free variable. However, in line (2) above, the variable that was bound to ꓯ does NOT become free. Rather, it gets bound to the ꓱ. This is illegal. The idea behind this should be intuitive. For instance, it does not follow from (1), Everything loves something that (2) Something loves itself. Note, however, that it IS permissible to choose a variable that is already bound (or in use ) by another quantifier so long as the variable being replaced does not become bound to some other quantifier. For instance, the following is permissible: 1 (1) (ꓯx)[Sxx (ꓱx)Lxx] A 1 (2) Sxx (ꓱx)Lxx 1, ꓯE This IS permissible since the x s in Lxx are bound by ꓱ, not ꓯ. For, ꓱ is the operator that occurs IMMEDIATELY to the left of Lxx. Here, the variables that were bound to the ꓯ (namely, the two x s in Sxx ) DO become free. In fact, the following are ALL permissible: 1 (1) (ꓯx)[Sxx (ꓱx)Lxx] A 1 (2) Sxx (ꓱx)Lxx 1, ꓯE 1 (3) Syy (ꓱx)Lxx 1, ꓯE 1 (4) Szz (ꓱx)Lxx 1, ꓯE When performing universal-elimination (ꓯE), you may ONLY make changes to the variables that were bound by the ꓯ being eliminated (i.e., we may only change the x s that occur in Sxx here). But, no matter what variable we pick when performing ꓯE here, the x s in (ꓱx)Lxx will ALWAYS remain unchanged, because those x s are bound by the ꓱ rather than the ꓯ. Lines (2) (4) are ALL permissible uses of ꓯE since, on these lines we did NOT alter the variables that were bound by the existential quantifier i.e., (ꓱx)Lxx 8

9 and furthermore, the variables that were bound to ꓯ on line (1) DID become free on lines (2) (4). Long story short, ꓯE now has the following restrictions: (1) You may ONLY choose/alter the variable that was bound to the ꓯ being eliminated. You may NOT alter any of the variables that are bound to any of the other quantifiers. (2) Whatever variable you choose, it must NOT appear freely in the assumption(s) that it rests on. (3) Whatever variable you choose, it MUST become free when performing ꓯE; that is, it must NOT become bound to some other quantifier. 5. ꓱE for Multi-Place Predicates: We will now add one final restriction to ꓱE: IF the Δ being stripped of its ꓱ falls within the scope of MORE THAN ONE operator, you may NOT pick a variable that is bound by one of those other operators unless it is the same variable that is bound by the ꓱ that is being eliminated. Confused? The example below will make things clearer. The following move is NOT permissible: 1 (1) (ꓱx)(ꓯy)Lxy A 1 (2) (ꓯy)Lyy Ass. (ꓱE) Clarification: It IS permissible, strictly speaking (you can assume whatever you want at any time), but this assumption cannot be used for the purposes of ꓱE. You cannot replace x with y here because y is already in use i.e., it is bound by ꓯ in the original statement that we are eliminating the ꓱ from. The idea behind this should be intuitive. For instance, it does not follow from (1), Something loves everything that (2) Everything loves itself. Note, however, that it IS permissible to choose a variable that is already bound by some other quantifier so long as the variable being replaced becomes free; i.e., does NOT become bound by another quantifier. For instance, the following move is permissible: 1 (1) (ꓱx)[Sxx (ꓯx)Lxx] A 1 (2) Sxx (ꓯx)Lxx Ass. (ꓱE) This is permissible because the x s in Lxx are bound by ꓯ, not ꓱ. What is important is that the x s in Sxx DO become free and DO NOT become bound by the ꓯ. So, the following moves are all permissible: 1 (1) (ꓱx)[Sxx (ꓯx)Lxx] A 1 (2) Sxx (ꓯx)Lxx Ass. (ꓱE) 1 (3) Syy (ꓯx)Lxx Ass. (ꓱE) 1 (4) Szz (ꓯx)Lxx Ass. (ꓱE) 9

10 When performing existential-elimination (ꓱE), your assumption for the purpose of ꓱE may ONLY make changes to the variables bound by the ꓱ being eliminated (i.e., we may only change the x s that occur in Sxx here). But, no matter what variable we pick when performing ꓱE here, the x s in (ꓯx)Lxx will ALWAYS remain unchanged. Lines (2) (4) are ALL suitable assumptions that we could make for the purpose of ꓱE. Note that on these lines we did NOT alter the variables that were bound by the universal quantifier i.e., (ꓯx)Lxx. Furthermore, the variables that were bound to the ꓱ DID become free in our ꓱE assumption; i.e., they did NOT become bound by the ꓯ. Long story short, ꓱE now has the following restrictions: (1) When making your ꓱE assumption, you may ONLY choose/alter the variable that was bound to the ꓱ that is removed. You may NOT alter any of the variables that are bound to any of the other quantifiers. (2) Whatever variable you choose, it must NOT appear freely in ANY prior assumption. (3) Whatever variable you choose, it MUST become free when making your assumption for the purposes of ꓱE; that is, it must NOT become bound to some other quantifier. (4) Also, the target of your ꓱE (i.e., the line where you finally perform ꓱE) must not contain any unbound instances of the variable that you chose to BECOME free in your ꓱE assumption. 10