Philosophy 110
We don't have a regular classroom video for this chapter, but see the PowerPoint video linked on the Main Page.
Let's start with the take away message.
Our language, common sense, and reasoning are much more complex than can be captured with just propositional logic (C9-C10). Although propositional logic is the foundation for both mathematics and logical reasoning, the beginning of the chapter shows its limitations.
Logicians have dealt with this situation by creating "higher" logics. Chapter 11 is an introduction to the first traditional extension of propositional logic, called quantification logic.
Given our time constraints, we will focus mostly on the translation basics and have just a brief overview of the proof process. We will also be skipping the Venn Diagram section. For the final exam you should be able to translate quantified statements and be able to do a simple proof in quantification logic.
Recall these examples from one of the tutorials for C1.
Example 1
Anyone who lives in the city Honolulu, HI also lives on the island of Oahu.
Kanoe lives in the city of Honolulu.
Therefore, Kanoe lives on the island of Oahu.
Example 2
Anyone who lives in the city Honolulu, HI also lives on the island of Oahu.
Kanoe does not live on the island of Oahu.
Therefore, Kanoe does not live in the city Honolulu, HI.
In the Why Logicians Don't Think Too Much section of C1, symbolic patterns were introduced similar to those below.
Anyone who lives in the city Honolulu, HI also lives on the island of Oahu.
Kanoe lives in the city of Honolulu.
Therefore, Kanoe lives on the island of Oahu.
For any x, if x is H, then x is O.
x is H.
So, x is O.
Then simplified:
If A, then B
A
So, B
Anyone who lives in the city Honolulu, HI also lives on the island of Oahu.
Kanoe does not live on the island of Oahu.
Therefore, Kanoe does not live in the city Honolulu, HI.
For any x, if x is H, then x is O.
x is not O.
So, x is not H.
Then simplified:
If A, then B.
Not B
So, not A
At this stage in the semester you should recognize the patterns of MP (Eg. 1) and MT (Eg. 2).
However, as an intro chapter we were actually cheating a little. Although good enough approximations, the actual patterns would be:
Example 1
For any x, if x is H, then x is O.
k is H.
So, k is O.
Example 2
For any x, if x is H, then x is O.
k is not O.
So, k is not H.
See why? The second premise and conclusion are not about any person. They refer to Kanoe.
The key point is that after reading C11 and studying the dictionary, you should see that now we will translate these arguments as follows:
1. (x)(Hx ⊃ Ox)
2. Hk / ∴ Ok
1. (x)(Hx ⊃ Ox)
2. ~Ok / ∴ ~Hk
The chapter notes some terminology.
(x) = a variable similar to p, q, r in C9 and C10.
Hx, Ox = propositional functions, empty places waiting for something to be substituted for the variable x.
k = an individual constant, a real thing or person.
Hk, Ok = substitution instances or instantiations of Hx and Ox.
There are four basic categorical propositions:
A (Universal Affirmative) -- All cigarette smokers are health risks.
(x)(Cx ⊃ Hx) --Given any x, if x is a C, then x is an H.
E (Universal Negative) -- No cigarette smoker is a health risk.
(x)(Cx ⊃ ~Hx) -- Given any x, if x is a C, then x is not an H.
I (Particular Affirmative) -- Some cigarette smokers are heath risks.
(∃x)(Cx • Hx) -- Some C's are H's.
O (Particular Negative) -- Some cigarette smokers are not health risks.
(∃x)(Cx • ~Hx) -- Some C's are not H's.
There are four new basic rules to add to our 19 rules from Chapters 9 and 10. Examples below presented a little less abstractly than covered in C11.
Universal Instantiation (UI)
(x)(Cx ⊃ Hx) / ∴ Ck ⊃ Hk
Justification: If any person who smokes cigarettes is a health risk, then if Kaiao smokes cigarettes, she is a health risk. "k" is a substitution instance or an instantiation of the variable x. We use letters a-w for substitution instances in proofs.
Universal Generalization (UG)
Cy ⊃ Hy / ∴ (x)(Cx ⊃ Hx)
Justification: If any arbitrarily selected individual (y) who smokes is a health risk, then anyone who smokes is a health risk.
This inference may seem to be a hasty conclusion, but it is not because of the meaning of the "y" special symbol. The only way this symbol can be inferred is from another universal generalization. We want to be able to prove simple arguments like this:
All people who smoke cigarettes are health risks.
All people who are health risks have high medical insurance payments.
So, all people who smoke cigarettes have high medical insurance payments.
1. (x)(Cx ⊃ Hx)
2. (x)(Hx ⊃ Ix) / ∴ (x)(Cx ⊃ Ix)
3. Cy ⊃ Hy ------(1) UI
4. Hy ⊃ Iy -------(2) UI
5. Cy ⊃ Iy -------(3)(4) HS
6. (x)(Cx ⊃ Ix) --(5) UG
Existential Generalization (EG)
Ck • Hk / ∴ (∃x)(Cx • Hx)
Justification: If Kaiao smokes cigarettes and she is a health risk, then there is at least one person (some) that smokes cigarettes and is a health risk.
Existential Instantiation (EI)
(∃x)(Cx • Hx) / ∴ Cμ • Hμ
Justification: If some people who smoke cigarettes are health risks, then there is at least one person who smokes cigarettes and is a health risk.
Notice the important restrictions for this rule.
If we did not have these restrictions, we could prove:
1. Some cigarette smokers are health risks.
2. Some birds are yellow.
So, some cigarette smokers are birds. [Obviously Invalid!]
For the change of quantifier rules, all you need to remember is this statement from the textbook:
"In using the change of quantifier rules mechanically, all you need to remember is this: If you see (x), you can change to ~(∃x)~ and vice versa; if you see (∃x), you can change to ~(x)~ and vice versa; if you see ~(x) you can change to (∃x)~ and vice versa; if you see ~(∃x) you can change to (x)~ and vice versa."
So:
(x) <--> ~(∃x)~
(x)(Cx ⊃ Hx)<-->~(∃x)~(Cx ⊃ Hx) or ~(∃x)(Cx • ~Hx)
~(x)~ <--> (∃x)
~(x)~(Cx ⊃ Hx)<-->(∃x)(Cx ⊃ Hx)
~(x) <--> (∃x)~
~(x)(Cx ⊃ Hx)<-->(∃x)~(Cx ⊃ Hx) or (∃x)(Cx • ~Hx)
~(∃x) <--> (x)~
~(∃x)(Cx • Hx) <--> (x)~(Cx • Hx) or (x)(Cx ⊃ ~Hx)
Let's see a few proof examples.
1. (x)[Jx ⊃ (Px • Tx)]
2. (∃x)(Dx • ~Tx) / ∴ (∃x)(Dx • ~Jx)
3. Da • ~Ta ----------(2)EI
4. Ja ⊃ (Pa • Ta) ---(1) UI
5. Da -----------------(3) Simp.
6. ~Ta • Da ----------(3) Com
7. ~Ta ----------------(6) Simp.
8. ~Ta v ~Pa --------(7) Add
9. ~Pa v ~Ta --------(8) Com
10. ~(Pa • Ta) -------(9) DeM
11. ~Ja ---------------(4)(10) MT
12. Da • ~Ja ---------(5)(11) Conj.
13. (∃x)(Dx • ~Jx) ---(12) EG
1. (∃x)(Px • ~Jx)
2. (∃x)(Dx • ~Tx)
3. (x)[Px ⊃ (Tx v Cx)] / ∴ (∃x)(Px • Cx)
4. Pa • ~Ja -----------(1) EI
5. Pa ⊃ (Ta v Ca) ---(3) UI
6. Pa ------------------(4) Simp.
7. Ta v Ca -----------(5)(6) MP
8. Da • ~Ta ----------(2) EI
9. ~Ta • Da ----------(8) Com.
10. ~Ta --------------(9) Simp.
11. Ca ---------------(7)(10) DS
12. Pa • Ca ---------(6)(11) Conj.
13. (∃x)(Px • Cx) --(12) EG
There is a mistake in Example 2. See it?
Try a few translations and proofs in Ex. VII. Suggested -- 2, 4, 5.