Several years ago I was teaching a logic course, and we were learning about formal proofs of validity. I enjoy proofs, and to keep myself sharp I was working through a practice quiz in David Kelley’s The Art of Reasoning, when I came across this argument:
D ⊃ (E ⊃ F)
D ⊃ (F ⊃ G)
∴ D ⊃ (E ⊃ G)
I was in a quiet library with plenty of time, but despite all my efforts I could not solve this (without using the Conditional Proof). The next day in class some students were finishing their assignment early, so I challenged them with this proof, thinking to myself, “That ought to keep them busy,” but not really expecting anyone to succeed. Before the end of class, Caroline Jones came forward and said, “I solved it, Mr. Nance.” I scoffed inwardly at first, only to be pleasantly surprised by her correct solution.
Since that time I have called this “The Caroline Jones” proof, and have challenged my logic students to solve it using only the regular rules of inference and replacement. The most elegant proof I have seen requires twelve total steps.
Anyone up to the challenge?
Hello and thank you for posting this, it really gave me a challenge and i enjoyed trying to solve it. I be exceedingly grateful if you could look at it and see if i got it right.
many thanks.
1)D ⊃ (E ⊃ F)
2)D ⊃ (F ⊃ G) / ∴ D ⊃ (E ⊃ G)
3)~D ⊃ ~(E ⊃ F) TRANS 1
4)~~D v ~(E ⊃ F) IMPL 3
5)D v ~(E ⊃ F) D. N. 4
6)~(E ⊃ F) v D COM 5
7)(E ⊃ F) ⊃ D IMPL 6
8)D ⊃ D H. S. 1, 7
9)~D v D IPML 8
10)~D ⊃ ~D TRANS 8
11)~~D v ~D IMPL 10
12)D v ~D D.N. 11
13)[(~D v D) • (D v ~D)] CONJ 12, 9
14)~D v (D • ~D) DIST 13
15)~D v (~D • D) COM 14
16)~D v ~D SIMP 15
17)~D TAUT 16
18)~D v ~E ADD 17,
19)~(D • E) DE M. 18
20)~(D • E) v G ADD 19
21)(D • E) ⊃ G IMPL 20
22)D ⊃ (E ⊃ G) EXP 21
This is a good attempt, but there are several errors that prevent it from working. Briefly:
Line 3 – Transposition switches the order, so it would give ~(E ⊃ F) ⊃ ~D
Line 14 – Distribution would take a little re-arranging, and would give ~D v (D • D) here.
Line 16 – Simplification is not valid within a larger compound proposition.
One hint that this proof goes in the wrong direction is that line 2 is never used, but the conclusion cannot be made from line 1 alone (as a shorter truth table would show). But you are on the right track a bit by trying to build the proof around Hypothetical Syllogism. Looking at the premises and the conclusion, you should see that you need to get F as the “middle term” so that it goes away in the H.S.
Blessings.
AGGGGGH please help me Mr. Nance. I have made it this far and i can’t seem to find another thing to do if you could check it over to see if I’m getting anything wrong v maybe give me a hint that would be much welcomed! Many thanks!
1)D ⊃ (E ⊃ F)
2)D ⊃ (F ⊃ G) / ∴ D ⊃ (E ⊃ G)
3)(D • E) ⊃ F EXP. 1
4)~F ⊃ ~(D • E) TRANS 3
5)F v ~(D • E) IMPL 4
6)(F v ~D) • (F v ~E) DIST 5
7)F v ~D SIMP. 6
8)~D v F COM 7
9)D ⊃ F IMPL 8
10)
Fewer errors this time, and your procedure is getting closer.
Line 5 – 4 Impl. gives ~~F v ~(D • E), then a D.N. to get yours.
Line 6 – Dist. is not allowed with the negation outside the (D • E).
Again, one hint that you have made a logical error is that D ⊃ F (from your step 9) is not a valid conclusion from D ⊃ (E ⊃ F).
I gave you this hint last time: Build the proof around Hypothetical Syllogism. Looking at the premises and the conclusion, you should see that you need to get F as the “middle term” so that it goes away in the H.S. So start like this:
1) D ⊃ (E ⊃ F)
2) D ⊃ (F ⊃ G) / ∴ D ⊃ (E ⊃ G)
3) (D • E) ⊃ F 1 Exp.
4) (D • F) ⊃ G 2 Exp.
After this, manipulate the proposition in step 4 until the F is the lone antecedent of a conditional. This allows you to connect it with the proposition in step 3 using H.S. After that, the proposition you have is equivalent to the desired conclusion, and it is only a matter of using the rules of replacement to manipulate it into the correct form.
Blessings!
Mr. Nance: Is there a way to get help with lesson 21, Intermediate Logic? Specifically column 7 and 10 of exercise 21. I have the answers, but I do not understand the steps leading to the answers. I have watched the DVD lesson about four times now, and I have completed all the other columns; however, I am still stumped on how to correctly arrive at the answers given for columns 7 and 10. Thank you.
Tom: Do you want me to give you all the steps, or would you like to work through them with me on a phone call or Skype? We could also use a Zoom classroom. Let me know; I am happy to help.
Wow! Fast reply! LOL, I ‘m at the library with two of my grandchildren, so maybe if you can email the steps, I can try to work on them when I get home. Thanks so much.
I will do that. It’s my pleasure.
Could you use this with a Conditional Proof?
3) D CPA
4) E ⊃ F 1,3, MP
5) F ⊃ G 2,3 MP
6) E ⊃ G 4,5 H.S.
7) D ⊃ (E⊃G) 3-6 C.P.
Can you check this? Thank you.
1) D⊃(E ⊃F)
2) D⊃(F ⊃G) / ∴ D ⊃(E ⊃G)
3) (D * E) ⊃F Ex. 1
4) (D * F) ⊃G EX. 2
5) (D * E) ⊃ [(D * E) * F] Abs. 3
6) (D * E) ⊃ [(E * D) * F) Comm. 5
7) (D * E) ⊃ [E * (D * F) Assoc. 6
8) (D * E) ⊃ (D * F) Simp. 7
9) ( D * E)⊃ G H.S. 4, 8
10) D ⊃ (E ⊃G) Ex. 9
Q.E.D