Today’s Lecture
4/20/10
8.5
--Intro to RAA Proofs
9.3
--Practice with Proofs
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 9.3 pg. 462 Part D (10-20 Even)
- Ex 9.3 pgs. 462-463 Part E (5-15 All)
- Read 9.4 464-469.
Announcements
CSUN Philosophy Student Conference
Fri., April 23, 2010
Whitsett Room, 4th Floor, Sierra Hall (CSUN campus)
8:30-2:30pm (followed by a barbecue at Northridge Park)
Program (available on Tu., April 20th)
More on the Final
-- Per the syllabus it’s cumulative and 40% of grade.
Chapter 1
--Be ready to define: valid, invalid, sound argument.
--Be ready to recognize whether arguments are sound, valid or invalid.
More on the Final
-- Per the syllabus it’s cumulative and 40% of grade.
Chapter 1
--Be ready to define: valid, invalid, sound argument.
--Be ready to recognize whether arguments are sound, valid or invalid.
Chapter 7:
--Be ready to define: counter-example, necessary condition, and
sufficient condition.
-- Symbolizations (study HW problems; memorize stylistic variants)
More on the Final
Chapter 8
--Direct proofs (study HW)
-- Conditional proofs (study HW)
More on the Final
Chapter 8
--Direct proofs (study HW)
-- Conditional proofs (study HW)
Chapter 9:
--Symbolizations (study HW)
-- Direct proofs (study HW)
-- RAA and Conditional Proofs (Study HW)
Intro to Reductio ad Absurdum (RAA) Proofs
RAA proofs, like conditional proofs, are a type of indirect
proof.
Intro to Reductio ad Absurdum (RAA) Proofs
RAA proofs, like conditional proofs, are a type of indirect
proof.
Constructing an RAA proof can make it easier to show an
argument to be valid.
Intro to Reductio ad Absurdum (RAA) Proofs
RAA proofs, like conditional proofs, are a type of indirect
proof.
Constructing an RAA proof can make it easier to show an
argument to be valid.
RAA proofs are based on the principle that:
Any statement that entails a contradiction (i.e. an
absurdity) is itself false.
Intro to Reductio ad Absurdum (RAA) Proofs
RAA proofs, like conditional proofs, are a type of indirect
proof.
Constructing an RAA proof can make it easier to show an
argument to be valid.
RAA proofs are based on the principle that:
Any statement that implies a contradiction (i.e. an
absurdity) is itself false.
Why think this principle is true? Well…
Intro to Reductio ad Absurdum (RAA) Proofs
According to the valid argument form Modus Tollens
(p! q, ~q " ~p), if a statement entails the truth of another
distinct statement, and the distinct statement for whatever
reason is false, then it must be the case that that the
original statement is false.
Intro to Reductio ad Absurdum (RAA) Proofs
Now let’s say that a statement entails that a contradiction (i.e. a
conjunction of a statement and its negation) is true. We could put this
symbolically as:
Intro to Reductio ad Absurdum (RAA) Proofs
Now let’s say that a statement entails that a contradiction (i.e. a
conjunction of a statement and its negation) is true. We could put this
symbolically as:
p ! (q • ~q)
Intro to Reductio ad Absurdum (RAA) Proofs
Now let’s say that a statement entails that a contradiction (i.e. a
conjunction of a statement and its negation) is true. We could put this
symbolically as:
p ! (q • ~q)
Well a contradiction by nature is always false (whatever the truth-value of
q is, the conjunction will always be false). This is just to say that the
negation of a contradiction will always be true. We can capture this fact
symbolically as:
Intro to Reductio ad Absurdum (RAA) Proofs
Now let’s say that a statement entails that a contradiction (i.e. a
conjunction of a statement and its negation) is true. We could put this
symbolically as:
p ! (q • ~q)
Well a contradiction by nature is always false (whatever the truth-value of
q is, the conjunction will always be false). This is just to say that the
negation of a contradiction will always be true. We can capture this fact
symbolically as:
~(q • ~q)
Intro to Reductio ad Absurdum (RAA) Proofs
Now let’s say that a statement entails that a contradiction (i.e. a
conjunction of a statement and its negation) is true. We could put this
symbolically as:
p ! (q • ~q)
Well a contradiction by nature is always false (whatever the truth-value of
q is, the conjunction will always be false). This is just to say that the
negation of a contradiction will always be true. We can capture this fact
symbolically as:
~(q • ~q)
Given MT, we can conclude ~p (i.e. p is false). Hence, any statement that
entails a contradiction is false.
Intro to Reductio ad Absurdum (RAA) Proofs
So, say an argument has the following conclusion: p. In
order to show p, we can assume ~p (on a line) and seek to
derive a contradiction: q ! ~q (on a line). The
contradiction shows that ~p must be false, which is to say
that p must be true!
If the conclusion is ~p, we can assume p and seek to derive
a contradiction. The contradiction shows that p must be
false, which is to say that ~p must be true.
Intro to Reductio ad Absurdum (RAA) Proofs
An Example:
1. (F # ~F) ! G
"G
Intro to Reductio ad Absurdum (RAA) Proofs
An Example:
1. (F # ~F) ! G
2. ~G
"G
Assume for RAA
Intro to Reductio ad Absurdum (RAA) Proofs
An Example:
1. (F # ~F) ! G
2. ~G
3. ~(F # ~F)
"G
Assume for RAA
1, 2 MT
Intro to Reductio ad Absurdum (RAA) Proofs
An Example:
1. (F # ~F) ! G
2. ~G
3. ~(F # ~F)
4. ~F • ~~F
"G
Assume for RAA
1, 2 MT
3, DeM
Intro to Reductio ad Absurdum (RAA) Proofs
An Example:
1. (F # ~F) ! G
2. ~G
3. ~(F # ~F)
4. ~F • ~~F
5. G
"G
Assume for RAA
1, 2 MT
3, DeM
2-4 RAA
Intro to Reductio ad Absurdum (RAA) Proofs
An Example:
1. (F # ~F) ! G
2. ~G
3. ~(F # ~F)
4. ~F • ~~F
5. G
"G
Assume for RAA
1, 2 MT
3, DeM
2-4 RAA
Intro to Reductio ad Absurdum (RAA) Proofs
An Example:
1. P ! (R • Q)
2. P ! (T • ~R)
" ~P
Intro to Reductio ad Absurdum (RAA) Proofs
An Example:
1. P ! (R • Q)
2. P ! (T • ~R)
" ~P
3. P
Assume for RAA
Intro to Reductio ad Absurdum (RAA) Proofs
An Example:
1. P ! (R • Q)
2. P ! (T • ~R)
" ~P
3. P
Assume for RAA
4. R • Q
1,3 MP
Intro to Reductio ad Absurdum (RAA) Proofs
An Example:
1. P ! (R • Q)
2. P ! (T • ~R)
" ~P
3. P
Assume for RAA
4. R • Q
1,3 MP
5. T • ~R
2,3 MP
Intro to Reductio ad Absurdum (RAA) Proofs
An Example:
1. P ! (R • Q)
2. P ! (T • ~R)
" ~P
3. P
Assume for RAA
4. R • Q
1,3 MP
5. T • ~R
2,3 MP
6. R
4, Simp
Intro to Reductio ad Absurdum (RAA) Proofs
An Example:
1. P ! (R • Q)
2. P ! (T • ~R)
" ~P
3. P
Assume for RAA
4. R • Q
1,3 MP
5. T • ~R
2,3 MP
6. R
4, Simp
7. ~R
5, Simp
Intro to Reductio ad Absurdum (RAA) Proofs
An Example:
1. P ! (R • Q)
2. P ! (T • ~R)
" ~P
3. P
Assume for RAA
4. R • Q
1,3 MP
5. T • ~R
2,3 MP
6. R
4, Simp
7. ~R
5, Simp
8. R • ~R
6,7 Conj
Intro to Reductio ad Absurdum (RAA) Proofs
An Example:
1. P ! (R • Q)
2. P ! (T • ~R)
" ~P
3. P
Assume for RAA
4. R • Q
1,3 MP
5. T • ~R
2,3 MP
6. R
4, Simp
7. ~R
5, Simp
8. R • ~R
6,7 Conj
9. ~P
3-8 RAA
Intro to Reductio ad Absurdum (RAA) Proofs
An Example:
1. P ! (R • Q)
2. P ! (T • ~R)
" ~P
3. P
Assume for RAA
4. R • Q
1,3 MP
5. T • ~R
2,3 MP
6. R
4, Simp
7. ~R
5, Simp
8. R • ~R
6,7 Conj
9. ~P
3-8 RAA
Answers to HW
We’ll come back to RAA on Thursday, for now let’s return
to our latest predicate logic homework.
Ex 9.3 pgs. 460-461 Part B (1-15 All).
#s 1-5
# 1. Line 2 is incorrect. (Can’t apply UI to a part of a line)
#2. Line 5 is incorrect. (Can’t univ. generalize a constant that is
introduced by EI). Line 8 is incorrect. (Can’t existentially
instantiate to a variable that occurs earlier in the proof)
#3. Line 2 is incorrect. (Can’t apply EI to only part of a line)
#4. Line 2 is incorrect. (Can’t simplify from a existentially
quant. Statement)
#5. Line 2 is incorrect. (Always instantiate to a constant)
#s 6-11
#6. No incorrect moves
#7. Line 2 is incorrect. (Need to instantiate to a constant)
#8. Line 2 is incorrect. (Can’t apply UI to only part of line)
#9. Line 3 is incorrect. (Can’t apply EG to only part of a line)
#10. Line 3 is incorrect. (Line 1 is not a conditional)
#11. Line 2 is incorrect. (x in line 1 not uniformly replaced).
Line 3 is incorrect. (line 2 not an instance of line 3)
#s 12-15
#12. Line 2 is incorrect. (x in line 1 not uniformly replaced)
#13. Line 3 is incorrect. (Line 2 is an instance of line 3.
But ‘a’ cannot appear in line 3)
#14. Line 5 is incorrect. (Line 4 is not an instance of line 5)
#15. Line 8 is incorrect. (Line 8 is a generalization of a
constant that appears in a line derived by EI).
More Answers to HW
Ex 9.3 pg. 461 Part C (1-8 All):
A Few General Tips
If the conclusion is an existentially quantified statement,
derive an instance of it and apply EG to the instance.
If the conclusion is a conditional (or contains a
conditional) -- and you’re not using CP -- you may need to
derive the conditional’s ‘corresponding’ disjunction and
then apply MI. The same strategy applies if the conclusion
is a disjunction (or contains a disjunction).
#1
1. (x)(Fx ! ~Gx)
2. Fa
" ($x)~Gx
#1
1. (x)(Fx ! ~Gx)
2. Fa
" ($x)~Gx
3. Fa ! ~Ga
1, UI
#1
1. (x)(Fx ! ~Gx)
2. Fa
" ($x)~Gx
3. Fa ! ~Ga
1, UI
4. ~Ga
2,3 MP
#1
1. (x)(Fx ! ~Gx)
2. Fa
" ($x)~Gx
3. Fa ! ~Ga
1, UI
4. ~Ga
2,3 MP
5. ($x)~Gx
4, EG
#2
1. Hc # Jd
2. (x)(Hx ! ($y)Ky)
3. (x)(Jx ! ($y)Ly)
" ($y)Ky # ($y)Ly
#2
1. Hc # Jd
2. (x)(Hx ! ($y)Ky)
3. (x)(Jx ! ($y)Ly)
4. Hc ! ($y)Ky
" ($y)Ky # ($y)Ly
2, UI
#2
1. Hc # Jd
2. (x)(Hx ! ($y)Ky)
3. (x)(Jx ! ($y)Ly)
4. Hc ! ($y)Ky
5. Jd ! ($y)Ly
" ($y)Ky # ($y)Ly
2, UI
3, UI
#2
1. Hc # Jd
2. (x)(Hx ! ($y)Ky)
3. (x)(Jx ! ($y)Ly)
4. Hc ! ($y)Ky
5. Jd ! ($y)Ly
6. ($y)Ky # ($y)Ly
" ($y)Ky # ($y)Ly
2, UI
3, UI
1,4,5 CD
#3
1. (x)(Mx ! Ox)
2. ~(Nc # ~Md)
" ($x)Ox
#3
1. (x)(Mx ! Ox)
2. ~(Nc # ~Md)
" ($x)Ox
3. ~Nc • ~~Md
2, DeM
#3
1. (x)(Mx ! Ox)
2. ~(Nc # ~Md)
" ($x)Ox
3. ~Nc • ~~Md
2, DeM
4. ~~Md
3, Simp
#3
1. (x)(Mx ! Ox)
2. ~(Nc # ~Md)
" ($x)Ox
3. ~Nc • ~~Md
2, DeM
4. ~~Md
3, Simp
5. Md
4, DN
#3
1. (x)(Mx ! Ox)
2. ~(Nc # ~Md)
3. ~Nc • ~~Md
4. ~~Md
5. Md
6. Md ! Od
" ($x)Ox
2, DeM
3, Simp
4, DN
1, UI
#3
1. (x)(Mx ! Ox)
2. ~(Nc # ~Md)
3. ~Nc • ~~Md
4. ~~Md
5. Md
6. Md ! Od
7. Od
" ($x)Ox
2, DeM
3, Simp
4, DN
1, UI
5,6 MP
#3
1. (x)(Mx ! Ox)
2. ~(Nc # ~Md)
3. ~Nc • ~~Md
4. ~~Md
5. Md
6. Md ! Od
7. Od
8. ($x)Ox
" ($x)Ox
2, DeM
3, Simp
4, DN
1, UI
5,6 MP
7, EG
#4
1. (z)Hz
2. (x)(Gx ! ~Hx)
" ($y)(Gy ! (x)Kx)
#4
1. (z)Hz
2. (x)(Gx ! ~Hx)
3. Ha
" ($y)(Gy ! (x)Kx)
1, UI
#4
1. (z)Hz
2. (x)(Gx ! ~Hx)
3. Ha
4. Ga ! ~Ha
" ($y)(Gy ! (x)Kx)
1, UI
2, UI
#4
1. (z)Hz
2. (x)(Gx ! ~Hx)
3. Ha
4. Ga ! ~Ha
5. ~~Ha
" ($y)(Gy ! (x)Kx)
1, UI
2, UI
3, DN
#4
1. (z)Hz
2. (x)(Gx ! ~Hx)
3. Ha
4. Ga ! ~Ha
5. ~~Ha
6. ~Ga
" ($y)(Gy ! (x)Kx)
1, UI
2, UI
3, DN
4,5 MT
#4
1. (z)Hz
2. (x)(Gx ! ~Hx)
3. Ha
4. Ga ! ~Ha
5. ~~Ha
6. ~Ga
7. ~Ga # (x)Kx
" ($y)(Gy ! (x)Kx)
1, UI
2, UI
3, DN
4,5 MT
6, Add
#4
1. (z)Hz
2. (x)(Gx ! ~Hx)
3. Ha
4. Ga ! ~Ha
5. ~~Ha
6. ~Ga
7. ~Ga # (x)Kx
8. Ga ! (x)Kx
" ($y)(Gy ! (x)Kx)
1, UI
2, UI
3, DN
4,5 MT
6, Add
7, MI
#4
1. (z)Hz
2. (x)(Gx ! ~Hx)
3. Ha
4. Ga ! ~Ha
5. ~~Ha
6. ~Ga
7. ~Ga # (x)Kx
8. Ga ! (x)Kx
9. ($y)(Gy ! (x)Kx)
" ($y)(Gy ! (x)Kx)
1, UI
2, UI
3, DN
4,5 MT
6, Add
7, MI
8, EG
#5
1. (x)(Fx ! Gx)
2. (x)(Hx ! Jx)
3. Fa # Ha
" (z)(Fz # Hz) ! ($x)(Gx # Jx)
#5
1. (x)(Fx ! Gx)
2. (x)(Hx ! Jx)
3. Fa # Ha
4. Fa ! Ga
" (z)(Fz # Hz) ! ($x)(Gx # Jx)
1, UI
#5
1. (x)(Fx ! Gx)
2. (x)(Hx ! Jx)
3. Fa # Ha
4. Fa ! Ga
5. Ha ! Ja
" (z)(Fz # Hz) ! ($x)(Gx # Jx)
1, UI
2, UI
#5
1. (x)(Fx ! Gx)
2. (x)(Hx ! Jx)
3. Fa # Ha
4. Fa ! Ga
5. Ha ! Ja
6. Ga # Ja
" (z)(Fz # Hz) ! ($x)(Gx # Jx)
1, UI
2, UI
3,4,5 CD
#5
1. (x)(Fx ! Gx)
2. (x)(Hx ! Jx)
3. Fa # Ha
4. Fa ! Ga
5. Ha ! Ja
6. Ga # Ja
7. ($x)(Gx # Jx)
" (z)(Fz # Hz) ! ($x)(Gx # Jx)
1, UI
2, UI
3,4,5 CD
6, EG
#5
1. (x)(Fx ! Gx)
2. (x)(Hx ! Jx)
3. Fa # Ha
" (z)(Fz # Hz) ! ($x)(Gx # Jx)
4. Fa ! Ga
1, UI
5. Ha ! Ja
2, UI
6. Ga # Ja
3,4,5 CD
7. ($x)(Gx # Jx)
6, EG
8. ~(z)(Fz # Hz) # ($x)(Gx # Jx) 7, Add
#5
1. (x)(Fx ! Gx)
2. (x)(Hx ! Jx)
3. Fa # Ha
" (z)(Fz # Hz) ! ($x)(Gx # Jx)
4. Fa ! Ga
1, UI
5. Ha ! Ja
2, UI
6. Ga # Ja
3,4,5 CD
7. ($x)(Gx # Jx)
6, EG
8. ~(z)(Fz # Hz) # ($x)(Gx # Jx) 7, Add
9. (z)(Fz # Hz) ! ($x)(Gx # Jx) 8, MI
#6
1. Ra ! Sa
2. (x)(Sx ! Tx)
" ($y)(Ry ! Ty)
3. Sa ! Ta
2, UI
4. Ra ! Ta
1, 3 HS
5. ($y)(Ry ! Ty) 4, EG
#7
1. (x)(y)[Lx ! (Mx • Ny)]
2. ~Mb
3. (y)[Lb ! (Mb • Ny)] 1, UI
4. Lb ! (Mb • Nb)
3, UI
5. ~Mb # ~Nb
2, Add
6. ~(Mb • Nb)
5, DeM
7. ~Lb
4,6 MT
8. ~Lb # Ob
7, Add
9. Lb ! Ob
8, MI
" Lb ! Ob
#8
1. ~~(v)Fv # ~(w)Gw
2. ~[($x)Hx # ~(w)Gw]
3. [(v)Fv • ~($x)Hx] ! ~(y)Jy
4. Ka ! (y)Jy
5. ~($x)Hx • ~~(w)Gw
6. ~~(w)Gw
7. ~~(v)Fv
8. (v)Fv
9. ~($x)Hx
10. (v)Fv • ~($x)Hx
11. ~(y)Jy
12. ~Ka
13. ~Ka # La
14. Ka ! La
15. ($z)(Kz ! Lz)
" ($z)(Kz ! Lz)
2, DeM
5, Simp
1,6 DS
7, DN
5, Simp
8,9 Conj
3, 10 MP
4, 11 MT
12, Add
13, MI
14, EG
Ex 9.3 pg. 462 Part D (10-20 Even):
Some More Tips
-- Always employ EI before UI
-- If the conclusion is a universally quantified statement,
derive an instance of it and apply UG to the instance.
#10
1. (y)(~Py ! ~Ly)
2. Lc # Ld
3. ~Pc ! ~Lc
4. Lc ! Pc
5. ~Pd ! ~Ld
6. Ld ! Pd
7. Pc # Pd
8. Pd # Pc
" Pd # Pc
1, UI
3, Cont
1, UI
5, Cont
2,4,6 CD
7, Com
#12
1. (z)[Uz ! (Kz # Sz)]
2. (z)Uz
3. ($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]
2. ~[~(x)Jx # ~Kc]
3. ~~(x)Jx • ~~Kc
4. (x)Jx • Kc
5. (x)Jx
6. Jc
7. Kc
8. Jc • Kc
" Jc • Kc
1, UI
2, DeM
3, DN DN
4, Simp
5, UI
4, Simp
6,7 Conj
#18
1. (x)[(Bx ! (z)Az) ! ~P]
2. ($x)~Bx
3. ~Ba
4. (Ba ! (z)Az) ! ~P
5. (~Ba # (z)Az) ! ~P
6. ~Ba # (z)Az
7. ~P
" ~P
2, EI
1, UI
4, MI
3, Add
5,6 MP
#20
1. (x)[(Sx • ~(z)Rz) ! Nd]
2. (x)~Nx
3. ($x)Sx
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
" Rc
Ex 9.3 pgs. 462-463 Part E (5-15 All)
#5
1. (x)(Jx ! ~Ex)
2. ( $x)(Jx # Jd)
3. Ja # Jd
4. Ja ! ~Ea
5. Jd ! ~Ed
6. ~Ea # ~Ed
7. Ea ! ~Ed
8. ($y)(Ey ! ~Ed)
" ($y)(Ey ! ~Ed)
2, EI
1, UI
1, UI
3,4,5 CD
6, MI
7, EG
#6
1. (x)(Lx ! Mx) ! (x)(Nx ! Lx)
2. (x)~Lx
3. ~La
2, UI
4. ~La # Ma
3, Add
5. La ! Ma
4, MI
6. (x)(Lx ! Mx)
5, UG
7. (x)(Nx ! Lx)
1, 6 MP
8. Na ! La
7, UI
9. ~Na
8,3 MT
10. (x)~Nx
9, UG
" (x)~Nx
#7
1. (x)(Sx ! Tx)
2. ($y)(Ry • ~Ty)
3. Ra • ~Ta
4. Sa ! Ta
5. ~Ta
6. ~Sa
7. Ra
8. Ra • ~Sa
9. ($z)(Rz • ~Sz)
" ($z)(Rz • ~Sz)
2, EI
1, UI
3, Simp
4,5 MT
3, Simp
7,6 Conj
8, EG
#8
1. (x)(Bx ! Cx)
2. (x)(Ax ! Bx)
3. (x)(Cx ! Dx)
4. ($x)~Dx
5. ~Da
6. Ca ! Da
7. Aa ! Ba
8. Ba ! Ca
9. ~Ca
10. ~Ba
11. ~Aa
12. ($x)~Ax
"($x)~Ax
4, EI
3, UI
2, UI
1, UI
6,5 MT
8,9 MT
7, 10 MT
11, EG
#9
1. (x)(Rx % Sx) " (x)(Rx ! Sx) • (x)(Sx ! Rx)
2. Ra % Sa
1, UI
3. (Ra ! Sa) • (Sa ! Ra)
2, ME
4. (Ra ! Sa)
3, Simp
5. (x)(Rx ! Sx)
4, UG
6. (Sa ! Ra)
3, Simp
7. (x)(Sx ! Rx)
6, UG
8. (x)(Rx ! Sx) • (x)(Sx ! Rx) 5,7 Conj
#10
1. (x)[(Bx # Ax) % Cx]
2. (x) ~Cx
3. (Ba # Aa) % Ca
4. [(Ba # Aa) ! Ca] • [Ca ! (Ba # Aa)]
5. ~Ca
6. (Ba # Aa) ! Ca
7. ~(Ba # Aa)
8. ~Ba • ~Aa
9. ~Aa
10. ~Aa # Ba
11. Aa ! Ba
12. ~Ba
13. ~Ba # Aa
14. Ba ! Aa
15. (Aa ! Ba) • (Ba ! Aa)
16. (Aa % Ba)
17. (x)(Ax % Bx)
" (x)(Ax % Bx)
1, UI
3, ME
2, UI
4, Simp
5,6 MT
7, DeM
8, Simp
9, Add
10, MI
8, Simp
12, Add
13, MI
11, 14 Conj
15, ME
16, UG
#11
1. (x)(Dx ! ~Kx)
2. ($x)(Ex • Hx)
3. (x)(Hx ! Dx)
4. (x)(Jx ! Kx)
5. Ea • Ha
6. Ha ! Da
7. Ha
8. Da
9. Da ! ~Ka
10. ~Ka
11. Ja !Ka
12. ~Ja
13. Ea
14. Ea • ~Ja
15. ($x)(Ex • ~Jx)
" ($x)(Ex • ~Jx)
2, EI
3, UI
5, Simp
6,7 MP
1, UI
8,9 MP
4, UI
10,11 MT
5, Simp
13, 12 Conj
14, EG
#12
1. (x)[Fx % (Hx • ~(y)Gy)]
2. ($x)~Fx
3. (z)Hz
4. ~Fa
5. Fa %(Ha • ~(y)Gy)
6. [Fa !(Ha • ~(y)Gy)] • [(Ha • ~(y)Gy) ! Fa]
7. (Ha • ~(y)Gy) ! Fa
8. ~(Ha • ~(y)Gy)
9. ~Ha # ~~(y)Gy
10. ~~Ha
11. (y)Gy
12. Gc
" Gc
2, EI
1,UI
5, ME
6, Simp
4,7 MT
8 DeM
3 UI, DN
9, 10 DS, DN
11, UI
#13
1. (x)[Bx ! (Cx • Dx)]
2. ($x)Bx
3. Ba
4. Ba ! (Ca • Da)
5. Ca • Da
6. ~~Ca • ~~Da
7. ~(~Ca # ~Da)
8. ($x)~(~Cx # ~Dx)
" ($x)~(~Cx # ~Dx)
2, EI
2, UI
3,4 MP
5, DN DN
6, DeM
7, EG
#14
1. (x)[Mx ! ($y)(Ny • Px)]
2. (x)(Nx ! ~G)
3. ($x)Mx
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
" ~G
#15
1. (x)[Rx ! (Sx # (y)Ty)]
2. (x)(Rx ! Sx) ! Pb
3. ~(y)Ty
4. Ra ! (Sa # (y)Ty)
5. ~Ra # (Sa # (y)Ty)
6. (~Ra # Sa) # (y)Ty
7. ~Ra # Sa
8. Ra ! Sa
9. (x)(Rx ! Sx)
10. Pb
" Pb
1, UI
4, MI
5, As
3,6 DS
7 MI
8, UG
2,9 MP