 Syllabus Update
 No class on Tuesday
 HW5 (part 2) now due on Wednesday
 No change in syllabus; we’ll just skip identity and functions
 Homework 5 (part 1)
 Stats: Mean=26.7/30, Max=30, Min=24.5
 Notes
 Always give variable order in your symbolization key
 Ex. L xy : x loves y
 Match your sentences to what is in your symbolization key; if a predicate takes two singular terms in the key, it must have 2 terms in the translations
 Symbolizing “everyone”
 We now have a way to use quantifiers to talk about every thing in the UD, but it is important to understand what that really means
 ‘  ’ means “no matter which x you choose…” – that means for every single thing in the UD
 ‘Charlie loves everyone’: (  x)Lcx
 Say your UD only consists of Charlie and Sue. This means we could pick either Charlie or Sue out of the bag. The point is, if
Charlie loves everyone, that includes himself (Lcc).
 Symbolization Techniques
 As with SL connectives, there are many different ways to say each kind of sentence in English. They still have the same form in PL regardless of the original sentence.
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 All dogs are furry. (All As are Bs.)
 Every dog is furry.
 Each dog is furry.
 Dogs are furry.
 A dog is furry.
 No dog is cold-blooded. (No As are Bs.)
 Dogs are not cold-blooded.
 A dog is not cold-blooded.
 Some dogs are yellow. (Some As are Bs.)
 There are yellow dogs.
 This is an existential claim. All it says is that, somewhere in the
UD bag o’ stuff, there is a yellow dog.
 Some dogs are not friendly. (Some As are not Bs.)
 There are unfriendly dogs.
 By now, you should feel pretty comfortable with these simple sentences
(the syllogisms) and be able to identify them even when they don’t show up in exactly the same way
 Of course, the sentences that appear in these templates can themselves be complex PL sentences
 Symbolizing sentences like this is where a paraphrase can be very helpful
 Ex. Everyone that Michael likes likes either Henry or Sue.
 1) What kind of sentence is it? Universal (  )
 2) Paraphrase (in steps): No matter what x you choose, if Michael likes x, then x likes either Henry or Sue.
 Get in the habit of hearing the word “if” right after you say “no matter what x I choose”. Almost all  sentences have a conditional as their immediate subformula!
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 Notice that I don’t immediately paraphrase the full consequent of the conditional. It is helpful to do your paraphrase in steps, especially if it contains more than one quantifier. This one doesn’t, so we could have skipped to a complete paraphrase.
 Note that the book uses a different paraphrase technique than the one presented here. (“Each thing is such that if Michael likes it, then either it likes Henry or it likes Sue.”)
 3) Complete Paraphrase (underline TF elements): No matter what x you choose, if Michael likes x, then either x likes Henry or x likes
Sue.
 4) Symbolization: (  x)(Lmx  (Lxh  Lxs))
 Notice that this turns out to be a classic “All As are Bs” sentence, but in this case, B is a disjunction.
 “Rita doesn’t like Michael but she likes everyone that Michael likes.”
 1) Sentence Type: Conjunction
 2) Paraphrase 1: Both it is not the case that Rita likes Michael and
Rita likes everyone that Michael likes.
 Paraphrase 2: Both it is not the case that Rita likes Michael and no matter what x I choose, if Michael likes x, then Rita likes x.
 Symbolization: ~Lrm & (  x)(Lmx  Lrx)
 “Pit bulls are dangerous but Labradors are not.”
 Conjunction
 Both pit bulls are dangerous and Labradors are not dangerous.
 Both no matter what x I choose, if x is a pit bull then x is dangerous and no matter what y I choose, if y is a Labrador then it is not the case that y is dangerous.
 Note that it would have been perfectly legal to use x as the only variable in this sentence because the second time it is used (for
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Labradors) it is not within the same scope as the first quantifier.
But it’s a good practice to use different variables even if you don’t have to
 (  x)(Px  Dx) & (  y)(Ly  ~Dy)
 “Pit bulls and Rottweilers are dangerous, but Labradors are not.”
 The basic structure of this sentence is the same as above. It is a conjunction and the right conjunct is the same, but the left conjunct is more complicated. Since we already know how to symbolize the right side, we can concentrate on the left side and then just add it to the conjunction
 “Pit bulls and Rottweilers are dangerous”
 Both pit bulls are dangerous and Rottweilers are dangerous
 We can give a shorter symbolization of this part of the sentence by making it only one 
 No matter what x I choose, if either x is a pit bull or x is a
Rottweiler, then x is dangerous.
 (  x)[(Px  Rx)  Dx]
 Notice that the antecedent must be a disjunction, not a conjunction. If we made it a conjunction, we would end up saying something about all things that are both pit bulls and
Rottweilers, which is not what we intend
 Complete Symbolization: (  x)[(Px  Rx)  Dx] & (  y)(Ly  ~Dy)
 “Every integer is either odd or even.” (UD: integers)
 This sentence is clearly a universal but also seems to be a disjunction. This is the exception to the rule about universals and conditionals – universals and disjunctions are allowed.
 Sentence Type: The main logical operator of this sentence must be
 , not  .
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 (  x)(Ox  Ex) – No matter what x I choose, either x is odd or x is even.
 (  x)Ox  (  y)Ey – Either every integer is odd or every integer is even. WRONG!
 “any”
 The quantity term “any” is used differently depending on context in
English. Sometimes it is a universal and sometimes it is an existential
 Universal ‘any’
 “Anyone who likes Sue likes Rita.” – (  x)(Lxs  Lxr)
 “Everyone who likes Sue likes Rita.” – (  x)(Lxs  Lxr)
 “If anyone likes Sue, he or she likes Rita.” – (  x)(Lxs  Lxr)
 Existential ‘any’
 “If anyone likes Sue, Michael does.” – (  x)Lxs  Lms
 This is an existential use of “any” because there is no pronominal
cross-reference. This means that there is no reference to the variable used in the antecedent of the English conditional in the consequent
 Rule of Thumb: When a quantity term is used in the antecedent of an English conditional and there is pronominal cross-reference to that quantity term in the consequent, then a universal quantifier is used
 Note that this symbolization is actually equivalent to (  x)(Lxs 
Lms), but more on that later
 Big Example (p. 322)
 1) All the cat’s-eyes belong to Rhoda.
 (  x)(Cx  Bxr)
 2) All the marbles but the shooters are cat’s-eyes.
 (  x)[(Mx & ~Sx)  Cx]
 3) Some, but not all, of the cat’s-eyes are green.
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 (  x)(Cx & Gx) & ~(  y)(Cy  Gy)
 4) None of the steelies is red, green, or blue.
 (  x)[Tx  ~(Rx  (Gx  Bx))]
 5) All of the shooters that are steelies belong to Terry.
 (  x)[(Sx & Tx)  Bxt]
 6) Some green marbles and some blue marbles but no red ones belong to Clarence.
 [(  x)((Mx & Gx) & Bxc) & (  y)((My & By) & Byc)] & ~(  z)((Mz &
Rz) & Bzc)
 7) Ashley wins all Clarence’s marbles.
 No matter what x you choose, if x is a marble that belongs to
Charles, then Ashley wins x.
 (  x)[(Mx & Bxc)  Wax]
 8) Rhoda wins all Terry’s Cat’s-eyes and shooters.
 No matter what x you choose, if x is either a Cat’s-eye or a shooter and x belongs to Terry, then Rhoda wins x.
 (  x)[((Cx  Sx) & Bxt)  Wrx]
 9) Terry doesn’t have any marbles.
 It is not the case that there is an x such that x is a marble and x belongs to Terry.
 ~(  x)(Mx & Bxt)
 10) Rhoda gives all the red marbles she wins to Clarence.
 No matter what x you choose, if x is a red marble that Rhoda won, then Rhoda gives x to Clarence.
 (  x)[((Mx & Rx) & Wrx)  Grxc]
 11) Clarence gives all his green marbles to Ashley and all his blue marbles to Terry.
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 No matter what x you choose: if x is a marble that belongs to
Clarence, then both if x is green then Clarence gives x to Ashley and if x is blue then Clarence gives x to Terry.
 (  x)[(Mx & Bxc)  ((Gx  Gcxa) & (Bx  Gcxt))]
 This could have been symbolized using two universal quantifiers, making it a conjunction instead of a universal.
 Multiple Quantifiers with Overlapping Scope
 So far, our symbolizations have generally consisted of just one quantifier, or perhaps two but never stacked. That is, when multiple quantifiers were used, they were spread out in a conjunction or other truth-functional sentence.
 Recall: “Everyone likes someone.”
 In this case, ‘likes’ is a two-place predicate, so we need two variables to fill the holes.
 Paraphrase: No matter what x you choose, there is some y such that x likes y – (  x)(  y)Lxy
 Contrast: “Someone is liked by everyone.”
 Paraphrase: There is some x such that no matter what y you choose, y likes x – (  x)(  y)Lyx
 Here we say that there is one particular person that everyone likes. In the first example, we only know that everyone likes someone, but it could be that everyone likes someone different or whatever.
 When the same type of quantifier is used, the order doesn’t matter
 “Everyone likes everyone” – (  x)(  y)Lxy
 “Someone likes someone” – (  x)(  y)Lxy
 Examples (Symbolization key: p. 335)
 “There is no largest positive integer.”
 No matter what x you choose, there is some y such that y is larger than x. – (  x)(  y)Lyx
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 “An odd number times an odd number is odd.”
 No matter what x and y you choose, if both x is odd and y is odd then x times y is odd. – (  x)(  y)((Dx &Dy)  Oxy)
 “No product of prime numbers is prime.”
 It is not the case that there is an x and there is a y such that x and y are prime numbers and x times y is prime. --
~(  x)(  y)((Px&Py)&Pxy)
 Way back to the Republican Example!
 None of David’s friends supports Republicans. Sarah supports Bush, and Bush is a Republican. So Sarah is no friend of David’s.
 Symbolization Key:
UD: People
Fxy: x is a friend of y
Sxy: x supports y
Rx: x is a Republican d: David s: Sarah b: Bush
 None of David’s friends supports Republicans.
 No matter what x you choose, if x is a friend of David’s then x does not support Republicans.
 x does not support Republicans – it is not the case that there is some y such that y is a Republican and x supports y.
 (  x)[Fxd  ~(  y)(Ry & Sxy)]
 x does not support Republicans – no matter what y you choose, if y is Republican then it is not the case that x supports y.
 (  x)[Fxd  (  y)(Ry  ~Sxy)]
 Sarah supports Bush, and Bush is a Republican.
 Ssb & Rb
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  Sarah is no friend of David’s.
 ~Fsd
 When to Stack
 You may be tempted to immediately stack all of the quantifiers you need at the beginning of your symbolization, but this is not always a good strategy
 Remember that the order matters if different types of quantifiers are used!
 I suggest only introducing quantifiers into your paraphrases (and therefore symbolizations) when you absolutely need to. That way, quantifiers will have their smallest possible scope
 If a quantifier only appears over part of a sentence that means that the variable it binds cannot appear in the other part of the sentence. If it did, it would be a free variable, and then the expression wouldn’t be a sentence.
 If you really would rather have the quantifiers at the start of your sentence, then there are several PL equivalence that allow for quantifier movement.
 Remember that these cases only deal with sentences that contain one part that has no occurrences of the variable in question.
 When this happens in a conjunction or a disjunction, you may extend the scope of the quantifier by simply moving it to the start of the sentence. The type of the quantifier does not change.
 When the quantifier only appears in the consequent of a conditional, the quantifier can be moved to the front of the sentence and it does not change.
 When the quantifier only appears in the antecedent of a conditional, the quantifier must be flipped and can be moved to the front of the sentence.
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 The scope of a quantifier on only one side of a biconditional cannot be changed.
 If you extend the scope of a quantifier so that it includes another quantifier, make sure that both aren’t binding the same variable!
 (  x)Px & (  x)Rx – this is a well-formed formula of PL. The x that appears on the right is not the same x that appears on the left. However, it is not well-formed to have quantifiers bind the same variable in a sentence. In order to make this sentence into a stacked version, one of the sides will need to be renamed.
 (  x)Px & (  y)Ry => (  x)(  y)(Px & Ry)
 Some of you will likely insist on stacking your quantifiers at the beginning of your symbolization. Just be aware that this is a dangerous practice, especially when different types of quantifiers are used.
 It is a good idea after you are done your symbolization to think about how it is paraphrased to see if it really does match the original sentence.
 Even more practice (Key: p. 342)
 Everyone who understands either Bertrand Russell’s Principia
Mathematica or Lewis Carroll’s Alice in Wonderland understands this text.
 No matter what x you choose, if (both x is a person and (either x understands PM or x understands AW)) then x understands this text.
 (  x)[(Px & (Uxp  Uxa))  Uxt]
 No one understands everything.
 No matter what x you choose, if x is a person then x does not understand everything.
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 It is not the case that no matter what y you choose, x understands y
 (  x)[Px  ~(  y)Uxy]
 No one understands anything.
 No matter what x you choose, if x is a person then x does not understand anything.
 It is not the case that there is some y such that x understands y.
 (  x)[Px  ~(  y)Uxy]
 If someone understands PM, then than person understands AW.
 Despite the presence of the word ‘someone’, this sentence has to be a universal. Remember to look for pronominal cross reference!
 No matter what x you choose, if x is a person that understands PM, then x understands AW.
 (  x)[(Px & Uxp)  Uxa]
 Only people who understand either PM or AW understand this text.
 This is akin to: People understand this text only if they understand either PM or AW.
 No matter what x you choose, if x is a person who understands this text then either x understands PM or x understands AW.
 (  x)[(Px & Uxt)  (Uxp  Uxa)]
 Anyone who understands anything is envied by someone.
 No matter what x you choose, if x is a person and x understands anything, then someone envies x.
 No matter what x you choose, if x is a person and there is some y such that x understands y, then there is some z such that z envies x.
 (  x)[(Px & (  y)Uxy)  (  z)(Pz &Ezx)]
 Anyone who understands everything is envied by everyone.
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 No matter what x you choose, if x is a person and no matter what y you choose, x understands y, then no matter what z you choose, if z is a person, then z envies x.
 (  x)[(Px & (  y)Uxy)  (  z)(Pz  Ezx)]
 Notice that the consequent ‘x is envied by everyone’ must be a universal over a conditional. If we tried to use the same symbolization as above and just replace (  z) with (  z), we would end up saying that everything is a person that envies x, which is not what was intended.
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