Today’s Lecture 4/15/10

9.1

Free/Bound Variables Cont.

9.3

4th Imp. Rule Cont. Practice w/ annotations

Announcements

Homework:

--Ex 9.3 pgs. 460-461 Part B (1-15 All).

--Ex 9.3 pg. 461 Part C (1- 8 All)

(per the directions only use UI, EG)

Exercises

9.1 pg. 431 Part B (1-15 all):

#’s 1-8

1. x in Hx is bound by (x); y in Gy is free.

2. x in Ax is bound by (x), y in By by (y), z is free

3. y in Cy bound by (!y)

4. both z’s are bound by (z); the x is free.

5. x in Kx bound by (!x); x in Lx bound by (x)

6. x in Mx bound by (x); y in ~Ny bound by the (y).

7. x and y in Pxy are bound by (x) and (y) respectively. The x in Qx is free.

8. y in Sy bound by (!y)

#’s 9-15

9. all of the x’s bound by (x)

10. x in Bx bound by (x); y in Dy bound by (y); z in Ez bound by (z)

11. x in Fxz bound by (x); z in Fxz bound by (!z); y in Fyz bound by (!y); the z in Fyz bound by (z).

12. z in Hz bound by (z); the y in Ky is free.

13. x in Lx bound by the first (!x); x in Nx bound by the second (!x)

14. x in Ox is free; x in Px bound by (x); y in Wy bound by (y)

15. z in Abz is free; y in By bound by (y); y in Cyx is free; x in Cyx bound by (x).

4th Rule: Universal Generalization (UG)

Consider the following:

All trees are plants. All plants are living things. So all trees are living things.

In symbolic form:

(x)(Tx " Px)

(x)(Px " Lx) # (x)(Tx " Lx)

4th Rule: Universal Generalization (UG)

In symbolic form:

(x)(Tx " Px)

(x)(Px " Lx) # (x)(Tx " Lx)

Intuitively, this is clearly valid. But how can we show that it’s valid by means of a proof?

What enables us to show this argument to be valid is the fact that we’re permitted to make a certain universal generalization (a certain claim about all things) under certain conditions. UG permits us to infer a universally quantified formula from an instance of a universally quantified formula, given certain conditions.

4th Rule: Universal Generalization (UG)

Consider the proof for our symbolic argument:

1. (x)(Tx " Px)

2. (x)(Px " Lx) # (x)(Tx " Lx)

3. Ta " Pa 1, UI

4. Pa " La 2, UI

5. Ta " La 3,4 HS

6. (x)(Tx " Lx) 5, UG

4th Rule: Universal Generalization (UG)

Line 6 is our universal generalization about all things. What justifies it?

We instantiated to the constant ‘a’ in 3 and 4, but nothing in the proof required us to instantiate to ‘a’. Given what 1 and 2 are claiming, we could have instantiated to any individual constant in 3 and 4, which would have resulted in any constant in 5. Because we could say of any constant (not just ‘a’) that if it were a tree, then it would be a living thing (line 5), we are permitted to claim that everything is such that if it is a tree, then it’s a living thing (line 6).

First Restriction on UG

Restriction 1: We cannot universally generalize from a constant that occurs in a premise in the argument to be proved.

First Restriction on UG

Say a premise in an argument says:

1. Ho

We can’t infer:

2. (x)Hx 1, incorrect use of UG

First Restriction on UG

Say a premise in an argument says:

1. Ho

We can’t infer:

2. (x)Hx 1, incorrect use of UG

If we didn’t have our first restriction, then we could say that 2 follows from 1. But we can come up with a substitution instance (see below) that shows that it’s not the case that 2 must be true (i.e. impossible to be false) given the truth of 1. Thus we can’t say that 2 logically follows from 1 via UG.

First Restriction on UG

Say a premise in an argument says:

1. Ho

We can’t infer:

2. (x)Hx 1, incorrect use of UG

If we didn’t have our first restriction, then we could say that 2 follows from 1 via UG. But we can come up with a substitution instance (see below) that shows that it’s not the case that 2 must be true (i.e. impossible to be false) given the truth of 1. Thus we can’t say that 2 logically follows from 1 via UG.

Obama is a human being T

Everything is such that it’s a human being F

First Restriction on UG

Consider:

1. Ms

2. (x)(Mx " Wx) # conclusion

3. Ms " Ws 2, UI

4. Ws 1,3 MP

5. ~Ws " ~Ms 3, Cont.

6. (x)(~Wx " ~Mx) 5, incorrect use of UG

First Restriction on UG

Consider:

1. Ms

2. (x)(Mx " Wx) # conclusion

3. Ms " Ws 2, UI

4. Ws 1,3 MP

5. ~Ws " ~Ms 3, Cont.

6. (x)(~Wx " ~Mx) 5, incorrect use of UG

While we universally generalized from 5, which is not a premise, it’s still the case that ‘s’ occurs in the premise. Thus it’s not a correct move.

First Restriction on UG

1. Ms

2. (x)(Mx " Wx) # conclusion

3. Ms " Ws 2, UI

4. Ws 1,3 MP

5. ~Ws " ~Ms 3, Cont.

6. (x)(~Wx " ~Mx) 5, incorrect use of UG

If the generalization of a constant that appears in a premise makes the generalized line logically follow the relevant line with the constant, then again obviously invalid inferences become valid. For example, if 6 follows from 5 by UG, then (x)Wx follows from line 4. In both cases we’re generalizing from a constant that appears in a premise. But

(x)Wx doesn’t follow 4. As evidenced by the following sub. instance:

First Restriction on UG

1. Simon Cowell is a mammal

2. Everything is such that if it’s a mammal, then it’s warm blooded

3. If Simon Cowell is a mammal, then he is warm-blooded 2, UI

4. Simon Cowell is warm-blooded 1,3 MP

5. Everything is such that it’s warm blooded. 4, UG

4 is true but 5 is false. This shows that (x)Wx doesn’t logically follow Ws in the previous symbolized argument.

Second Restriction on UG

Restriction 2: We cannot universally generalize from a constant that appears on a line derived from EI.

Second Restriction on UG

Say a proof has the following two lines:

1. (!x)Sx

2. Sd 1, EI

We cannot infer:

3. (x)Sx 2, incorrect use of UG

Second Restriction on UG

Say a proof has the following two lines:

1. (!x)Sx

2. Sd 1, EI

We cannot infer:

3. (x)Sx 2, incorrect use of UG

Without our second restriction, we could claim that 3 follows from 2 via UG. But say 1 claims: ‘something is such that it takes steroids; and say 2 claims: ‘the thing alluded to in 1, dub him John Doe, takes steroids. 3 would then say that everything is such that it takes steroids, which is false. So clearly 3 doesn’t follow 2.

An Important Tip

We can only make a universal generalization about constants that are introduced by the rule UI.

Third Restriction on UG

One must ensure that the constant that one generalizes from does not appear in the resulting universally quantified statement (if one wants to apply UG to a line, the constant in that line must not appear in the resulting universally quantified statement).

Third Restriction on UG

Example:

1. (x)(Mx " Px) # (x)(Mc " Px)

2. Mc " Pc 1, UI

3. (x)(Mc " Px) 2, incorrect use of UG

Third Restriction on UG

Example:

1. (x)(Mx " Px) # (x)(Mc " Px)

2. Mc " Pc 1, UI

3. (x)(Mc " Px) 2, incorrect use of UG

The constant c in line 2 appears in 3, the resulting universally quantified statement. Say 2 says: ‘if Curious George is a monkey, then Curious George is a primate. 3 would then say: everything is such that it is a primate (all things are primates) if Curious George is a Monkey. 2 is true but 3 is false (true antecedent and false consequent). 3 then, certainly doesn’t follow from 2.

Brief Summary of UG

A line in a proof follows from some previous line via UG if and only if the line that follows is a universally quantified statement, and the line from which it follows is an instance of it (the universally quantified statement), and where the constant in the instance does not occur in the premise of the argument, a previous line derived by EI, nor in the relevant universally quantified statement itself.

Answers to HW

Ex 9.3 Pgs. 458-460 Part A (1-15 All)

#1

1. (x)(Rx " Tx)

2. ~Tc # ~Rc

3. Rc " Tc 1, UI

4. ~Rc 2,3 MT

#2

1.Km

2. (!x)Kx " (x)Lx #Lm

3. (!x)Kx 1, EG

4. (x)Lx 2,3 MP

5. Lm 4, UI

#3

1. (z)(Az " Bz)

2.(!y)Ay # (!y)By

3. Ab 2, EI

4. Ab " Bb 1, UI

5. Bb 3,4 MP

6. (!y)By 5, EG

#4

1. Hn #(!x)Hx

2. (!x)Hx 1, EG

#5

1. (x)(Rx " ~Ox)

2. (!y)(Sy • Ry) # (!z)(Sz • ~Oz)

3. Sb • Rb 2, EI

4. Rb " ~Ob 1, UI

5. Rb 3, Simp

6. ~Ob 4,5 MP

7. Sb 3, Simp

8. Sb • ~Ob 6,7 Conj

9. (!z)(Sz • ~Oz) 8, EG

#6

1. (z)(Mz • Lz)

2. (z)Mz " Kd #(!y)Ky

3. Mc • Lc 1, UI

4. Mc 3, Simp

5. (z)Mz 4, UG

6. Kd 2,5 MP

7. (!y)Ky 6, EG

#7

1. (y)(Ry " Ny)

2. ~Ng # (!y)~Ry

3. Rg " Ng 1,UI

4. ~Rg 3,2 MT

5. (!y)~Ry 4, EG

#8

1. Ab $ (Bb • Cb)

2. (x)~Cx # (!x)Ax

3. (Ab $ Bb) • (Ab $ Cb) 1, Dist

4. Ab $ Cb 3, Simp

5. ~Cb 2, UI

6. Ab 4,5 DS

7. (!x)Ax 6, EG

#9

1. (x)(Fx " Gx)

2. (x)(Gx " Fx) # (x)(Fx %Gx)

3. Fa " Ga 1, UI

4. Ga " Fa 2,UI

5. (Fa "Ga) • (Ga "Fa) 3,4 Conj

6. Fa % Ga 5, ME

7. (x)(Fx % Gx) 6, UG

#10

1. (!y)Py " (z)(~Nz $ Oz)

2. Pn

3. ~Om # (!x)~Nx

4. (!y)Py 2, EG

5. (z)(~Nz $ Oz) 1,4 MP

6. ~Nm $ Om 5, UI

7. ~Nm 3,6 DS

8. (!x)~Nx 7, EG

#11

1. (x)(Kx " Lx)

2. (x)(Lx " Mx) # (x)(Kx " Mx)

3. Ka " La 1, UI

4. La " Ma 2, UI

5. Ka " Ma 3,4 HS

6. (x)(Kx " Mx) 5, UG

#12

1. (x)[(Ax $ Bx) " Cx]

2. (x)~Cx # (x)~Bx

3. (Aa $ Ba) " Ca 1,UI

4. ~Ca 2,UI

5. ~(Aa $ Ba) 3,4 MT

6. ~Aa • ~Ba 5, DeM

7. ~Ba 6, Simp

8. (x)~Bx 7, UG

#13

1. (y)(Dy • Ey) # (y)Dy • (y)Ey

2. Db • Eb 1, UI

3. Db 2, Simp

4. (y)Dy 3, UG

5. Eb 2, Simp

6. (y)Ey 5, UG

7. (y)Dy • (y)Ey 4,6 Conj

#14

1. (!x)(Jx • Kx) # (!x)Jx • (!x)Kx

2. Jb • Kb 1, EI

3. Jb 2, Simp

4. (!x)Jx 3, EG

5. Kb 2, Simp

6. (!x)Kx 5, EG

7. (!x)Jx • (!x)Kx 4,6 Conj.

#15

1. (x)[Ma $ (Lx • Nx)]

2. ~Ma #(x)Nx

3. Ma $ (Lb • Nb) 1, UI

4. Lb • Nb 2,3 DS

5. Nb 4, Simp

6. (x)Nx 5, UG

Ex 9.3 pgs. 460-461 Part B (1-15 All)