Predicate logic CP and IP (8.4):
Conditional proof (CP):
Remember from before that doing CP will give you your assumption horseshoed (as the main
connective) to where you leave the indented area.
There are two different ways to assume for a predicate logic CP, depending on whether the
horseshoe is the main connective of the conclusion, OR whether the quantifier is on the outside of
a horseshoe statement.
See the two proofs below:
A. 1. ( x)(Tx v ~Ox)
~( x)(Zx
B. 1. ( x)(Zx v Mx)
2. ( x)Mx
( x)(~Ox
/ (x)(~Zx
Mx)
Zx)
/ (x)~Zx
Ox)
( x)Ox
In the first proof, the (x) is acting like the “main connective” (governing the whole line) of the
conclusion, while in the second proof, the actual horseshoe is the main connective.
--When the is the actual main connective, you will assume the entire “p” or left side of it.
--When the quantifier is on the outside of a horseshoe statement, you will ignore it and only
assume the „p” or left side of the horseshoe statement within the parentheses/brackets. You will
put the quantifier back on after the CP ends.
A. 1. ( x)(Tx v ~Ox)
~( x)(Zx
2. ~Zx
3. ~Zx v Mx
4. Zx Mx
5. ( x)(Ax Mx)
6. ~( x)(Tx v ~Ox)
7. (x)~(Tx v ~Ox)
8. (x)(~Tx · Ox)
9. ~Tx · Ox
10. Ox
11. ~Zx Ox
12. (x)(~Zx Ox)
B. 1. ( x)(Zx v Mx)
2. ( x)Mx
( x)(~Ox
3. (x)~Zx
4. Za v Ma
5. ~Za
6. Ma
7. ( x)Mx
8. ( x)(~Ox Zx)
9. ~Ob Zb
10. ~Zb
11. Ob
12. ( x)Ox
13. (x)~Zx
( x)Ox
Zx)
Mx)
/ (x)(~Zx
ACP
2 Add
3 Impl.
4 EG
1,5 MT
6 CQ
7 DM
8 UI
9 simp.
2-10 CP
11 UG
/ (x)~Zx
ACP
1 EI
3 UI
4,5 DS
6 EG
2,7 MP
8 EI
3 UI
9,10 MT
11 EG
3-12 CP
( x)Ox
Ox)
There is a NEW UG restriction for CP and IP. Make sure you know the textbook
wording for the restrictions for the multiple-choice part of the final exam. The
restriction says that IF your assumption is a FREE VARIABLE (like in
proof A), you cannot UG that variable within that indented area . So in
proof A we could not have UG-ed anything with an “x”. We had to do the UG after we
left the indented area. This is only a UG restriction. You can still EG anything. This
new UG restriction would not have been an issue with Proof B, because we didn‟t assume
a free variable. A free variable is a variable (like Ax, Yx v Dx etc) that does NOT have a
quantifier attached to it.
Multi-assumption CP:
1. (x)(Jx · Zx) v (x)(Ox Mx)
2. ( x)Mx
(x)Ax
/ ( x)~Jx
(x)(Ox
3. ( x)~Jx
ACP
4. Ox
ACP
5. ~Ja
3 EI
6. ~Ja v ~Za
5 Add
7. ~(Ja · Za)
6 DM
8. ( x)~(Jx · Zx)
7 EG
9. ~(x)(Jx · Zx)
8 CQ
10. (x)(Ox Mx)
1,9 DS
11. Ox Mx
10 UI
12. Mx
4,11 MP
13. ( x)Mx
12 EG
14. (x)Ax
2,13 MP
15. Ax
14 UI
16. Ox Ax
4-15 CP
17. (x)(Ox Ax)
16 UG
18. ( x)~Jx
(x)(Ox Ax)
3-17 CP
Ax)
Triplebar CP is the same as before.
Indirect Proof (IP):
The new restriction rarely comes up with IP because you are almost always making an
assumption with a quantifier on it. IP is pretty much done exactly the way it was done before.
You assume the OPPOSITE of the whole conclusion, break stuff down until you find two
contradictory statements, conjoin them, then step back out with the OPPOSITE of your
assumption.
1. (x)Tx
2. ( x)(Tx · ~Ox)
(x)Zx
/ ( x)Ox v (x)(Tx · Zx)
3. ~[ ( x)Ox v (x)(Tx · Zx) ]
AIP
4. ~( x)Ox · ~(x)(Tx · Zx)
3 DM
5. ~( x)Ox
4 simp.
6. ~(x)(Tx · Zx)
4 simp.
7. (x)~Ox
5 CQ
8. ( x)~(Tx · Zx)
6 CQ
9. ( x)(~Tx v ~Zx)
8 DM
10. ~Ta v ~Za
9 EI
11. Ta
12. ~Za
13. ( x)~Zx
14. ~(x)Zx
15. ~( x)(Tx · ~Ox)
16. (x)~(Tx · ~Ox)
17. (x)(~Tx v Ox)
18. ~Ta v Oa
19. Oa
20 ( x)Ox
21. ( x)Ox · ~( x)Ox
22. ( x)Ox v (x)(Tx · Zx)
1 UI
10,11 DS
12 EG
13 CQ
2,14 MT
15 CQ
16 DM
17 UI
11,18 DS
19 EG
5,20 conj.
3-21 IP
Don’t forget that the opposite of a statement is just one tilde in the very front of (left
of) that statement.
A Triplebar CP
1. (x)Jx ≡ (∃x)(Kx v Mx)
2. (x)~Mx
3. ~Oa v (x)Rx
4. (x)(Jx v Ax) ⊃ (x)Ox
/ (x)(Jx  Ox) ≡ [ (∃x)Kx  (x)Rx ]
5. [(x)Jx ⊃ (∃x)(Kx v Mx)]  [(∃x)(Kx v Mx) ⊃ (x)Jx]
1 equiv
6. (x)Jx ⊃ (∃x)(Kx v Mx)
5 simp.
7. (∃x)(Kx v Mx) ⊃ (x)Jx
5 simp.
8. (x)(Jx  Ox)
ACP
9. Jx  Ox
8 UI
10. Jx
9 simp.
11. Jx v Ax
10 Add
12. (x)(Jx v Ax)
11 UG
13. (x)Ox
4,11 MP
14. Oa
13 UI
15. (x)Rx
3,14 DS
16. (x)Jx
10 UG
17. (∃x)(Kx v Mx)
6,16 MP
18. Kb v Mb
17 EI
19. ~Mb
2 UI
20. Kb
18, 19 DS
21. (∃x)Kx
20 EG
22. (∃x)Kx  (x)Rx
15,21 conj.
23. (x)(Jx  Ox) ⊃ [ (∃x)Kx  (x)Rx ]
8-22 CP
24. (∃x)Kx  (x)Rx
ACP
25. (∃x)Kx
24 simp.
26. Kc
25 EI
27. Kc v Mc
26 Add
28. (∃x)(Kx v Mx)
27 EG
29. (x)Jx
7,28 MP
30. Jx
29 UI
31. Jx v Ax
30 Add
32. (x)(Jx v Ax)
31 UG
33. (x)Ox
4,32 MP
34. Ox
33 UI
35. Jx  Ox
30,34 conj.
36. (x)(Jx  Ox)
35 UG
37. [ (∃x)Kx  (x)Rx ] ⊃ (x)(Jx  Ox)
24-36 CP
38. {(x)(Jx  Ox) ⊃ [ (∃x)Kx  (x)Rx ]}  {[ (∃x)Kx  (x)Rx ]⊃ (x)(Jx  Ox)} 23,37 conj.
39. (x)(Jx  Ox) ≡ [ (∃x)Kx  (x)Rx ]
38 Equiv.