TodayÕs Lecture 4/22/10 - CSUN

Today’s Lecture 4/22/10

9.3

--More Practice with Direct Proofs for Pred Logic

9.4

--Intro to QN, RAA, CP for Pred Logic

Announcements

-- Final Exam on May 11th (now’s the time to start

studying)!

-- Next Tues is the deadline to turn in any late homework

-- Homework:

- Ex 9.4 pg. 469 Part A (1-10 All, 14, 15, 16).

[note that for #4 – the answer is in the back of the book – there is no need for a DN move. The last edition of our text required this, but it has since expanded the QN rule. In short, the DN

annotation is a mistake].

Announcements

Recommendation for the Homework:

-For problems 1-5, 8, and 16 use direct proof

-For problems 6,7,9, 14 use conditional proof.

-For problem 10 use RAA; for problem 15 use conditional proof but also use RAA to derive the consequent.

Ex 9.3 pg. 462 Part D (10-20 Even):

Some More Tips

-- Always employ EI before UI

-- If the conclusion is a universally quantified statement, derive an instance of it and apply UG to the instance.

#10

1. (y)(~Py ! ~Ly)

2. Lc " Ld # Pd " Pc

3. ~Pc ! ~Lc 1, UI

4. Lc ! Pc 3, Cont

5. ~Pd ! ~Ld 1, UI

6. Ld ! Pd 5, Cont

7. Pc " Pd 2,4,6 CD

8. Pd " Pc 7, Com

#12

1. (z)[Uz ! (Kz " Sz)]

2. (z)Uz

3. ($z)~Sz # ($z)Kz

4. ~Sa 3, EI

5. Ua ! (Ka " Sa) 1, UI

6. Ua 2, UI

7. Ka " Sa 5,6 MP

8. Ka 7,4 DS

9. ($z)Kz 8, EG

#14

1. (x)[Cx ! (Dx • ($y)Ey)]

2. ~Db # ($x)~Cx

3. Cb ! (Db • ($y)Ey) 1, UI

4. ~Db " ~($y)Ey 2, Add

5. ~(Db • ($y)Ey) 4, DeM

6. ~Cb 3,5 MT

7. ($x)~Cx 6, EG

#16

1. (z)~[~(x)Jx " ~Kz] # Jc • Kc

2. ~[~(x)Jx " ~Kc] 1, UI

3. ~~(x)Jx • ~~Kc 2, DeM

4. (x)Jx • Kc 3, DN DN

5. (x)Jx 4, Simp

6. Jc 5, UI

7. Kc 4, Simp

8. Jc • Kc 6,7 Conj

#18

1. (x)[(Bx ! (z)Az) ! ~P]

2. ($x)~Bx # ~P

3. ~Ba 2, EI

4. (Ba ! (z)Az) ! ~P 1, UI

5. (~Ba " (z)Az) ! ~P 4, MI

6. ~Ba " (z)Az 3, Add

7. ~P 5,6 MP

#20

1. (x)[(Sx • ~(z)Rz) ! Nd]

2. (x)~Nx

3. ($x)Sx # Rc

4. Sa 3, EI

5. (Sa • ~(z)Rz) ! Nd 1, UI

6. ~Nd 2, UI

7. ~(Sa • ~(z)Rz) 5,6 MT

8. ~Sa " ~~(z)Rz 7, DeM

9. ~~Sa 4, DN

10. ~~(z)Rz 8,9 DS

11. (z)Rz 10, DN

12. Rc 11, UI

Ex 9.3 pgs. 462-463 Part E (5-15 All)

#5

1. (x)(Jx ! ~Ex)

2. ( $x)(Jx " Jd) # ($y)(Ey ! ~Ed)

3. Ja " Jd 2, EI

4. Ja ! ~Ea 1, UI

5. Jd ! ~Ed 1, UI

6. ~Ea " ~Ed 3,4,5 CD

7. Ea ! ~Ed 6, MI

8. ($y)(Ey ! ~Ed) 7, EG

#6

1. (x)(Lx ! Mx) ! (x)(Nx ! Lx)

2. (x)~Lx # (x)~Nx

3. ~La 2, UI

4. ~La " Ma 3, Add

5. La ! Ma 4, MI

6. (x)(Lx ! Mx) 5, UG

7. (x)(Nx ! Lx) 1, 6 MP

8. Na ! La 7, UI

9. ~Na 8,3 MT

10. (x)~Nx 9, UG

#7

1. (x)(Sx ! Tx)

2. ($y)(Ry • ~Ty) # ($z)(Rz • ~Sz)

3. Ra • ~Ta 2, EI

4. Sa ! Ta 1, UI

5. ~Ta 3, Simp

6. ~Sa 4,5 MT

7. Ra 3, Simp

8. Ra • ~Sa 7,6 Conj

9. ($z)(Rz • ~Sz) 8, EG

#8

1. (x)(Bx ! Cx)

2. (x)(Ax ! Bx)

3. (x)(Cx ! Dx)

4. ($x)~Dx #($x)~Ax

5. ~Da 4, EI

6. Ca ! Da 3, UI

7. Aa ! Ba 2, UI

8. Ba ! Ca 1, UI

9. ~Ca 6,5 MT

10. ~Ba 8,9 MT

11. ~Aa 7, 10 MT

12. ($x)~Ax 11, EG

#9

1. (x)(Rx % Sx) # (x)(Rx ! Sx) • (x)(Sx ! Rx)

2. Ra % Sa 1, UI

3. (Ra ! Sa) • (Sa ! Ra) 2, ME

4. (Ra ! Sa) 3, Simp

5. (x)(Rx ! Sx) 4, UG

6. (Sa ! Ra) 3, Simp

7. (x)(Sx ! Rx) 6, UG

8. (x)(Rx ! Sx) • (x)(Sx ! Rx) 5,7 Conj

#10

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

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

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

4. [(Ba " Aa) ! Ca] • [Ca ! (Ba " Aa)] 3, ME

5. ~Ca 2, UI

6. (Ba " Aa) ! Ca 4, Simp

7. ~(Ba " Aa) 5,6 MT

8. ~Ba • ~Aa 7, DeM

9. ~Aa 8, Simp

10. ~Aa " Ba 9, Add

11. Aa ! Ba 10, MI

12. ~Ba 8, Simp

13. ~Ba " Aa 12, Add

14. Ba ! Aa 13, MI

15. (Aa ! Ba) • (Ba ! Aa) 11, 14 Conj

16. (Aa % Ba) 15, ME

17. (x)(Ax % Bx) 16, UG

#11

1. (x)(Dx ! ~Kx)

2. ($x)(Ex • Hx)

3. (x)(Hx ! Dx)

4. (x)(Jx ! Kx) # ($x)(Ex • ~Jx)

5. Ea • Ha 2, EI

6. Ha ! Da 3, UI

7. Ha 5, Simp

8. Da 6,7 MP

9. Da ! ~Ka 1, UI

10. ~Ka 8,9 MP

11. Ja !Ka 4, UI

12. ~Ja 10,11 MT

13. Ea 5, Simp

14. Ea • ~Ja 13, 12 Conj

15. ($x)(Ex • ~Jx) 14, EG

#12

1. (x)[Fx % (Hx • ~(y)Gy)]

2. ($x)~Fx

3. (z)Hz # Gc

4. ~Fa 2, EI

5. Fa %(Ha • ~(y)Gy) 1,UI

6. [Fa !(Ha • ~(y)Gy)] • [(Ha • ~(y)Gy) ! Fa] 5, ME

7. (Ha • ~(y)Gy) ! Fa 6, Simp

8. ~(Ha • ~(y)Gy) 4,7 MT

9. ~Ha " ~~(y)Gy 8 DeM

10. ~~Ha 3 UI, DN

11. (y)Gy 9, 10 DS, DN

12. Gc 11, UI

#13

1. (x)[Bx ! (Cx • Dx)]

2. ($x)Bx # ($x)~(~Cx " ~Dx)

3. Ba 2, EI

4. Ba ! (Ca • Da) 2, UI

5. Ca • Da 3,4 MP

6. ~~Ca • ~~Da 5, DN DN

7. ~(~Ca " ~Da) 6, DeM

8. ($x)~(~Cx " ~Dx) 7, EG

#14

1. (x)[Mx ! ($y)(Ny • Px)]

2. (x)(Nx ! ~G)

3. ($x)Mx # ~G

4. Ma 3, EI

5. Ma ! ($y)(Ny • Pa) I, UI

6. ($y)(Ny • Pa) 4,5 MP

7. Nb • Pa 6, EI

8. Nb 7, Simp

9. Nb ! ~G 2, UI

10. ~G 8,9 MP

#15

1. (x)[Rx ! (Sx " (y)Ty)]

2. (x)(Rx ! Sx) ! Pb

3. ~(y)Ty # Pb

4. Ra ! (Sa " (y)Ty) 1, UI

5. ~Ra " (Sa " (y)Ty) 4, MI

6. (~Ra " Sa) " (y)Ty 5, As

7. ~Ra " Sa 3,6 DS

8. Ra ! Sa 7 MI

9. (x)(Rx ! Sx) 8, UG

10. Pb 2,9 MP

An Equivalence Rule: QN

Consider:

“Something is red”. Symbolically, we can say, ($x)Rx.


Consider:


This is equivalent to saying, “ it’s not the case that everything is not red”. Symbolically: ~(x)~Rx.


Consider:


This is equivalent to saying, “ it’s not the case that everything is not red”. Symbolically: ~(x)~Rx.

[Everything is not red? No. For something is red].


Consider:

“Something is not red”. ($x)~Rx.


Consider:

“Something is not red”. ($x)~Rx.

This is equivalent to saying, “it’s not the case that everything is red”. ~(x)Rx


Consider:

“Everything is red”. (x)Rx.


Consider:


This is like saying, “it’s not the case that something is not red. ~($x)~Rx


Consider:


This is like saying, “it’s not the case that something is not red. ~($x)~Rx

[Something is not red? No. For everything is red].


Consider:

“Everything is not red”. (x)~Rx.


Consider:

“Everything is not red”. (x)~Rx.

This is like saying, “it’s not the case that something is red”. ~($x)Rx


The statements in each aforementioned pair logically imply one another. If one is true, the other must be true. They are, after all, just different ways of saying the same thing.


The statements in each aforementioned pair logically imply one another. If one is true, the other must be true. They are, after all, just different ways of saying the same thing.

These statements illustrate the equivalence rule of Quantifier Negation (QN). Below are the four different forms:

($x)P is equiv. to ~(x)~P

($x)~P is equiv to ~(x)P

(x)P is equiv to ~($x)~P

(x)~P is equiv. to ~($x)P


QN can be applied to parts of lines. So we can have inferences such as the following:

1. (x)~Hx ! (y)Ky

2. ~($x)Hx ! (y)Ky 1, QN

1. (y)By ! ~(x)Ax

2. (y)By ! ($x)~Ax 1, QN

Another Tip

As a general rule, you may find it helpful to apply QN to a line that contains a negated quantifier [e.g. ~($x), ~(x)]. This can set you up for a UI or EI move.

Practice Proof -- Direct

1. ~(x)(Ax ! Mx)

2. (x)(Rx ! Mx) # ($x)(Ax • ~Rx)


1. ~(x)(Ax ! Mx)

2. (x)(Rx ! Mx) # ($x)(Ax • ~Rx)

3. ($x)~(Ax ! Mx) 1, QN


1. ~(x)(Ax ! Mx)

2. (x)(Rx ! Mx) # ($x)(Ax • ~Rx)

3. ($x)~(Ax ! Mx) 1, QN

4. ~(Aa ! Ma) 3, EI


1. ~(x)(Ax ! Mx)

2. (x)(Rx ! Mx) # ($x)(Ax • ~Rx)

3. ($x)~(Ax ! Mx) 1, QN

4. ~(Aa ! Ma) 3, EI

5. ~(~Aa " Ma) 4, MI


1. ~(x)(Ax ! Mx)

2. (x)(Rx ! Mx) # ($x)(Ax • ~Rx)

3. ($x)~(Ax ! Mx) 1, QN

4. ~(Aa ! Ma) 3, EI

5. ~(~Aa " Ma) 4, MI

6. ~~Aa • ~Ma 5, DeM


1. ~(x)(Ax ! Mx)

2. (x)(Rx ! Mx) # ($x)(Ax • ~Rx)

3. ($x)~(Ax ! Mx) 1, QN

4. ~(Aa ! Ma) 3, EI

5. ~(~Aa " Ma) 4, MI

6. ~~Aa • ~Ma 5, DeM

7. ~Ma 6, Simp


1. ~(x)(Ax ! Mx)

2. (x)(Rx ! Mx) # ($x)(Ax • ~Rx)

3. ($x)~(Ax ! Mx) 1, QN

4. ~(Aa ! Ma) 3, EI

5. ~(~Aa " Ma) 4, MI

6. ~~Aa • ~Ma 5, DeM

7. ~Ma 6, Simp

8. Ra ! Ma 2, UI


1. ~(x)(Ax ! Mx)

2. (x)(Rx ! Mx) # ($x)(Ax • ~Rx)

3. ($x)~(Ax ! Mx) 1, QN

4. ~(Aa ! Ma) 3, EI

5. ~(~Aa " Ma) 4, MI

6. ~~Aa • ~Ma 5, DeM

7. ~Ma 6, Simp

8. Ra ! Ma 2, UI

9. ~Ra 7,8 MT


1. ~(x)(Ax ! Mx)

2. (x)(Rx ! Mx) # ($x)(Ax • ~Rx)

3. ($x)~(Ax ! Mx) 1, QN

4. ~(Aa ! Ma) 3, EI

5. ~(~Aa " Ma) 4, MI

6. ~~Aa • ~Ma 5, DeM

7. ~Ma 6, Simp

8. Ra ! Ma 2, UI

9. ~Ra 7,8 MT

10. ~~Aa 6, Simp


1. ~(x)(Ax ! Mx)

2. (x)(Rx ! Mx) # ($x)(Ax • ~Rx)

3. ($x)~(Ax ! Mx) 1, QN

4. ~(Aa ! Ma) 3, EI

5. ~(~Aa " Ma) 4, MI

6. ~~Aa • ~Ma 5, DeM

7. ~Ma 6, Simp

8. Ra ! Ma 2, UI

9. ~Ra 7,8 MT

10. ~~Aa 6, Simp

11. Aa 10, DN


1. ~(x)(Ax ! Mx)

2. (x)(Rx ! Mx) # ($x)(Ax • ~Rx)

3. ($x)~(Ax ! Mx) 1, QN

4. ~(Aa ! Ma) 3, EI

5. ~(~Aa " Ma) 4, MI

6. ~~Aa • ~Ma 5, DeM

7. ~Ma 6, Simp

8. Ra ! Ma 2, UI

9. ~Ra 7,8 MT

10. ~~Aa 6, Simp

11. Aa 10, DN

12. Aa • ~Ra 11, 9 Conj


1. ~(x)(Ax ! Mx)

2. (x)(Rx ! Mx) # ($x)(Ax • ~Rx)

3. ($x)~(Ax ! Mx) 1, QN

4. ~(Aa ! Ma) 3, EI

5. ~(~Aa " Ma) 4, MI

6. ~~Aa • ~Ma 5, DeM

7. ~Ma 6, Simp

8. Ra ! Ma 2, UI

9. ~Ra 7,8 MT

10. ~~Aa 6, Simp

11. Aa 10, DN

12. Aa • ~Ra 11, 9 Conj

13. ($x)(Ax • ~Rx) 12 EG

RAA Proofs for Predicate Logic

Note well that with RAA proofs (and conditional proofs), we cannot universally generalize from a constant that appears in an unboxed assumption.

If we were permitted to do this, then it becomes possible to show that clearly invalid arguments are valid. But it’s impossible to show that an invalid argument is valid (any proof that shows an invalid argument to be valid is mistaken). Thus we’re not permitted to univ. generalize from a constant that appears in an unboxed assumption.

There is no such problematic consequence though if we universally generalize from a constant that appears in a boxed assumption.

Universal Generalization (UG)

A Brief Summary of Universal Generalization (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 a premise of the argument, a previous line derived by EI, in the universally quantified statement itself, nor in an unboxed (un-discharged) assumption.

Another Tip

As a general rule, it is oftentimes helpful to construct an RAA proof when the conclusion to be proved is an existentially quantified statement.


Use an RAA proof to show the following argument to be valid:

1. (x)(Px ! Sx)

2. Pa " Pb # ($x)Sx



1. (x)(Px ! Sx)

2. Pa " Pb # ($x)Sx

3. ~($x)Sx Assume for RAA



1. (x)(Px ! Sx)

2. Pa " Pb # ($x)Sx


4. (x)~Sx 3, QN



1. (x)(Px ! Sx)

2. Pa " Pb # ($x)Sx


4. (x)~Sx 3, QN

5. Pa ! Sa 1, UI



1. (x)(Px ! Sx)

2. Pa " Pb # ($x)Sx


4. (x)~Sx 3, QN

5. Pa ! Sa 1, UI

6. ~Sa 4, UI



1. (x)(Px ! Sx)

2. Pa " Pb # ($x)Sx


4. (x)~Sx 3, QN

5. Pa ! Sa 1, UI

6. ~Sa 4, UI

7. ~Pa 5,6MT



1. (x)(Px ! Sx)

2. Pa " Pb # ($x)Sx


4. (x)~Sx 3, QN

5. Pa ! Sa 1, UI

6. ~Sa 4, UI

7. ~Pa 5,6MT

8. Pb 2,7 DS



1. (x)(Px ! Sx)

2. Pa " Pb # ($x)Sx


4. (x)~Sx 3, QN

5. Pa ! Sa 1, UI

6. ~Sa 4, UI

7. ~Pa 5,6MT

8. Pb 2,7 DS

9. Pb ! Sb 1, UI



1. (x)(Px ! Sx)

2. Pa " Pb # ($x)Sx


4. (x)~Sx 3, QN

5. Pa ! Sa 1, UI

6. ~Sa 4, UI

7. ~Pa 5,6MT

8. Pb 2,7 DS

9. Pb ! Sb 1, UI

10. Sb 8,9 MP



1. (x)(Px ! Sx)

2. Pa " Pb # ($x)Sx


4. (x)~Sx 3, QN

5. Pa ! Sa 1, UI

6. ~Sa 4, UI

7. ~Pa 5,6MT

8. Pb 2,7 DS

9. Pb ! Sb 1, UI

10. Sb 8,9 MP

11. ~Sb 4, UI



1. (x)(Px ! Sx)

2. Pa " Pb # ($x)Sx


4. (x)~Sx 3, QN

5. Pa ! Sa 1, UI

6. ~Sa 4, UI

7. ~Pa 5,6MT

8. Pb 2,7 DS

9. Pb ! Sb 1, UI

10. Sb 8,9 MP

11. ~Sb 4, UI

12. Sb • ~Sb 10, 11 Conj



1. (x)(Px ! Sx)

2. Pa " Pb # ($x)Sx


4. (x)~Sx 3, QN

5. Pa ! Sa 1, UI

6. ~Sa 4, UI

7. ~Pa 5,6MT

8. Pb 2,7 DS

9. Pb ! Sb 1, UI

10. Sb 8,9 MP

11. ~Sb 4, UI

12. Sb • ~Sb 10, 11 Conj

13. ($x)Sx 3-12 RAA



1. (x)(Px ! Sx)

2. Pa " Pb # ($x)Sx


4. (x)~Sx 3, QN

5. Pa ! Sa 1, UI

6. ~Sa 4, UI

7. ~Pa 5,6MT

8. Pb 2,7 DS

9. Pb ! Sb 1, UI

10. Sb 8,9 MP

11. ~Sb 4, UI

12. Sb • ~Sb 10, 11 Conj

13. ($x)Sx 3-12 RAA

Another Tip

If the conclusion is a universally quantified statement that contains a conditional, use CP to derive the conditional and then apply UG

CP for Predicate Logic

1. (x)(Fx ! Gx)

2. (x)(Fx ! Hx) # (x)[Fx ! (Gx • Hx)]


1. (x)(Fx ! Gx)

2. (x)(Fx ! Hx) # (x)[Fx ! (Gx • Hx)]

3. Fa Assume for CP


1. (x)(Fx ! Gx)

2. (x)(Fx ! Hx) # (x)[Fx ! (Gx • Hx)]

3. Fa Assume for CP

4. Fa ! Ga 1, UI


1. (x)(Fx ! Gx)

2. (x)(Fx ! Hx) # (x)[Fx ! (Gx • Hx)]

3. Fa Assume for CP

4. Fa ! Ga 1, UI

5. Fa ! Ha 2, UI


1. (x)(Fx ! Gx)

2. (x)(Fx ! Hx) # (x)[Fx ! (Gx • Hx)]

3. Fa Assume for CP

4. Fa ! Ga 1, UI

5. Fa ! Ha 2, UI

6. Ga 3,4 MP


1. (x)(Fx ! Gx)

2. (x)(Fx ! Hx) # (x)[Fx ! (Gx • Hx)]

3. Fa Assume for CP

4. Fa ! Ga 1, UI

5. Fa ! Ha 2, UI

6. Ga 3,4 MP

7. Ha 3,5 MP


1. (x)(Fx ! Gx)

2. (x)(Fx ! Hx) # (x)[Fx ! (Gx • Hx)]

3. Fa Assume for CP

4. Fa ! Ga 1, UI

5. Fa ! Ha 2, UI

6. Ga 3,4 MP

7. Ha 3,5 MP

8. Ga • Ha 6,7 Conj


1. (x)(Fx ! Gx)

2. (x)(Fx ! Hx) # (x)[Fx ! (Gx • Hx)]

3. Fa Assume for CP

4. Fa ! Ga 1, UI

5. Fa ! Ha 2, UI

6. Ga 3,4 MP

7. Ha 3,5 MP


9. Fa ! (Ga • Ha) 3-8 CP


1. (x)(Fx ! Gx)

2. (x)(Fx ! Hx) # (x)[Fx ! (Gx • Hx)]

3. Fa Assume for CP

4. Fa ! Ga 1, UI

5. Fa ! Ha 2, UI

6. Ga 3,4 MP

7. Ha 3,5 MP


9. Fa ! (Ga • Ha) 3-8 CP


1. (x)(Fx ! Gx)

2. (x)(Fx ! Hx) # (x)[Fx ! (Gx • Hx)]

3. Fa Assume for CP

4. Fa ! Ga 1, UI

5. Fa ! Ha 2, UI

6. Ga 3,4 MP

7. Ha 3,5 MP


9. Fa ! (Ga • Ha) 3-8 CP

10. (x)[Fx ! (Gx • Hx)] 9, UG

Ex 9.4 pg. 469-470 Part A (1-10 All, 14, 15, 16).

-For problems 1-5, 8, and 16 use direct proof

-For problems 6,7,9, 14 use conditional proof.

-For problem 10 use RAA; for problem 15 use conditional proof but also use RAA to derive the consequent.

Possible answers are on following slides:

#1

1. (x)Ax ! (x)Bx

2. ~(x)Bx # ($x)~Ax

3. ~(x)Ax 1,2 MT

4. ($x)~Ax 3, QN

#2

1. ~($y)Cy

2. (y)~Cy ! (z)Dz # Db

3. (y)~Cy 1, QN

4. (z)Dz 2,3 MP

5. Db 4, UI

#3

1. ~(x)~Fx # ($x)Fx

2. ($x)Fx 1, QN

#4

1. ~($x)~Gx #(x)Gx

2. (x)Gx 1, QN

#5

1. ($y)Hy ! ($y)Jy

2. (y)~Jy # ~Ha

3. ~($y)Jy 2, QN

4. ~($y)Hy 1,3 MT

5. (y)~Hy 4, QN

6. ~Ha 5, UI

#6

1. (z)[(Kz " Lz) ! Mz] # (z)(Lz ! Mz)

2. La Assume for CP

3. (Ka " La) ! Ma 1, UI

4. Ka " La 2, Add

5. Ma 3,4 MP

6. La ! Ma 2-5 CP

7. (z)(Lz ! Mz) 6, UG

#7

1. (x)(Nx ! Ox) # ~(x)Ox ! ~(x)Nx

2. ~(x)Ox Assume CP

3. ($x)~Ox 2, QN

4. ~Oa 3, EI

5. Na ! Oa 1, UI

6. ~Na 5,4 MT

7. ($x)~Nx 6, EG

8. ~(x)Nx 7, QN

9. ~(x)Ox ! ~(x)Nx 2-8 CP

#8

1. ~($x)~Px

2. ~($y)Sy " ~(x)Px # ~Sd

3. (x)Px 1, QN

4. ~~(x)Px 3, DN

5. ~($y)Sy 2,4 DS

6. (y)~Sy 5, QN

7. ~Sd 6, UI

#9

1. (x)~Rx ! ($x)~~Tx # (x)~Tx ! ($x)Rx

2. (x)~Tx Assume CP

3. ~($x)Tx 2, QN

4. (x)~Rx ! ($x)Tx 1, DN

5.~(x)~Rx 3,4 MT

6. ($x)Rx 5, QN

7. (x)~Tx ! ($x)Rx 2-6 CP

#10

1. (x)(Ax ! ~Bx)

2. (y)Ay # (z)~Bz

3. ~(z)~Bz Assume RAA

4. ($z)Bz 3, QN

5. Ba 4, EI

6. Aa ! ~Ba 6, UI

7. Aa 2, UI

8. ~Ba 6,7 MP

9. Ba • ~Ba 5,8 Conj

10. (z)~Bz 3-9 RAA

#14

1. ($x)[Fx • (y)(Gy ! Hx)]

2. (x)[Fx ! (y)(By ! ~Hx)] # (x)(Gx ! ~Bx)

3. Ga Assume CP

4. Fb • (y)(Gy ! Hb) 1, EI

5. Fb 4, Simp

6. (y)(Gy ! Hb) 4, Simp

7. Fb ! (y)(By ! ~Hb) 2, UI

8. (y)(By ! ~Hb) 5, 7 MP

9. Ga ! Hb 6, UI

10. Hb 3,9 MP

11. Ba ! ~Hb 8, UI

12. ~~Hb 10, DN

13. ~Ba 11, 12 MT

14. Ga ! ~Ba 3-13 CP

15. (x)(Gx ! ~Bx) 14, UG

#15

1. (x)[Dx ! ($y)(Fy • Gy)] # (x)~Fx ! ~($y)Dy

2. (x)~Fx Assume CP

3. ($y)Dy Assume RAA

4. Da 3, EI

5. Da ! ($y)(Fy • Gy) 1, UI

6. ($y)(Fy • Gy) 4,5 MP

7. Fb • Gb 6, EI

8. ~Fb 2, UI

9. Fb 7, Simp

10. ~Fb • Fb 8,9 Conj

11. ~($y)Dy 3-10 RAA

12. (x)~Fx ! ~($y)Dy 2-11 CP

#16

1. ~(x)Mx " ($x)~Mx

2. ($x)Sx ! (x)Mx

3. Sb " (x)~Px # ~Pa

4. ~(x)Mx " ~(x)Mx 1, QN

5. ~(x)Mx 4, Re

6. ~($x)Sx 2,5 MT

7. (x)~Sx 6, QN

8. ~Sb 7, UI

9. (x)~Px 3,8 DS

10. ~Pa 9, UI

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TodayÕs Lecture 4/22/10 - CSUN