Easy question on Logic and Modes Ponens

Question

I got confused with these:

using ONLY this three axioms and Modus Ponens:$$1. \ F \implies (G\implies F) \\ 2. \ (F \implies (G\implies H))\implies ((F \implies G)\implies (F \implies H)) \\ 3. \ (\neg G \implies \neg F)\implies ((\neg G\implies F)\implies G)$$

I need to prove: $$4. \ (F \implies G)\implies ((G \implies H)\implies(F\implies H))\\$$ I see it intuitively, but I have to use ONLY the axioms.

I started like this: $$1. F \implies G\\2. \ H \implies (F \implies G) \ by \ (1)\\3.((H\implies F)\implies (H \implies G)) \ by \ (2)$$ but that is not what I want

Thank you for your help.

Do you want a formal proof or is it enough to prove that $$\vdash (F \to G)\to ((G \to H)\to(F\to H))?$$ Have you learned the deduction theorem yet? — Git Gud, Dec 13 '14 at 02:23
@GitGud yes, it is enough. I can't use it, that's the point. I am supposed to use only the three axioms above. — tmac_balla, Dec 13 '14 at 02:24
You can use the deduction theorem to get a formal proof, that's the point of the theorem. Have you learned about it? Did you understand its proof? — Git Gud, Dec 13 '14 at 02:27
You're contradicting yourself for the second time. There are two ways to read the question. One is "Provide a formal proof of $(F \to G)\to ((G \to H)\to(F\to H))$" (which corresponds to using only the axioms) the other is "Prove that $\vdash (F \to G)\to ((G \to H)\to(F\to H))$ (however you like)". I believe you want a formal proof. — Git Gud, Dec 13 '14 at 02:33
@GitGud Oh, ok. I need a formal one then, using the deduction thm. — tmac_balla, Dec 13 '14 at 02:43
You can't use the deduction theorem in a formal proof, when a formal proof is defined as sequence of formulas which starts with the axioms and ends with the theorem and every step is either an axiom or is provable from the axioms of modus ponens and substitution alone. — Doug Spoonwood, Dec 13 '14 at 05:49
Here you can find the proof of the "derived rule" ; $A→B,B→C⊢A→C$. With Deduction Th it is straightforward to derive the tautology; without it, we have to follow the "instruction" provided by the proof of the Deduction Th. — Mauro ALLEGRANZA, Dec 14 '14 at 16:07
Here you can find the first two "steps" in the proof of $⊢(F→G)→((G→H)→(F→H))$ without DT, i.e. the proof of $(F→G)⊢(G→H)→(F→H)$. — Mauro ALLEGRANZA, Dec 15 '14 at 20:47
See the answer to this post for a proof without deduction Theorem. — Mauro ALLEGRANZA, Dec 17 '14 at 11:02

score 0 · Answer 1 · answered Dec 13 '14 at 06:11

The automated theorem prover OTTER has found a 10 step, level 5 proof:

----> UNIT CONFLICT at   4.26 sec ----> 11611 [binary,11610.1,2.1] $F.

Length of proof is 10.  Level of proof is 5.

---------------- PROOF ----------------

1 [] -P(i(x,y))| -P(x)|P(y).
2 [] -P(i(i(a,b),i(i(b,c),i(a,c)))).
3 [] P(i(x,i(y,x))).
4 [] P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))).
5 [hyper,1,4,4] P(i(i(i(x,i(y,z)),i(x,y)),i(i(x,i(y,z)),i(x,z)))).
7 [hyper,1,3,4] P(i(x,i(i(y,i(z,u)),i(i(y,z),i(y,u))))).
8 [hyper,1,3,3] P(i(x,i(y,i(z,y)))).
16 [hyper,1,4,7] P(i(i(x,i(y,i(z,u))),i(x,i(i(y,z),i(y,u))))).
18 [hyper,1,5,8] P(i(i(x,i(i(y,x),z)),i(x,z))).
42 [hyper,1,16,16] P(i(i(x,i(y,i(z,u))),i(i(x,i(y,z)),i(x,i(y,u))))).
56 [hyper,1,3,18] P(i(x,i(i(y,i(i(z,y),u)),i(y,u)))).
230 [hyper,1,42,3] P(i(i(i(x,y),i(z,x)),i(i(x,y),i(z,y)))).
431 [hyper,1,18,56] P(i(i(i(x,y),z),i(y,z))).
11610 [hyper,1,431,230] P(i(i(x,y),i(i(y,z),i(x,z)))).
11611 [binary,11610.1,2.1] $F.

Interestingly enough, there exists at least two proofs with a greater level, but which have less steps. OTTER found a 9 step, level 7 proof:

 ----> UNIT CONFLICT at   0.17 sec ----> 514 [binary,513.1,2.1] $F.

 Length of proof is 9.  Level of proof is 7.

 ---------------- PROOF ----------------

 1 [] -P(i(x,y))| -P(x)|P(y).
 2 [] -P(i(i(a,b),i(i(b,c),i(a,c)))).
 3 [] P(i(x,i(y,x))).
 4 [] P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))).
 5 [hyper,1,4,4] P(i(i(i(x,i(y,z)),i(x,y)),i(i(x,i(y,z)),i(x,z)))).
 8 [hyper,1,3,3] P(i(x,i(y,i(z,y)))).
 14 [hyper,1,5,8] P(i(i(x,i(i(y,x),z)),i(x,z))).
 24 [hyper,1,3,14] P(i(x,i(i(y,i(i(z,y),u)),i(y,u)))).
 56 [hyper,1,14,24] P(i(i(i(x,y),z),i(y,z))).
 120 [hyper,1,3,56] P(i(x,i(i(i(y,z),u),i(z,u)))).
 125 [hyper,1,56,5] P(i(i(x,y),i(i(x,i(y,z)),i(x,z)))).
 275 [hyper,1,4,120] P(i(i(x,i(i(y,z),u)),i(x,i(z,u)))).
 513 [hyper,1,275,125] P(i(i(x,y),i(i(y,z),i(x,z)))).
 514 [binary,513.1,2.1] $F.

And a 7 step, level 6 proof:

 ----> UNIT CONFLICT at   0.37 sec ----> 2045 [binary,2044.1,2.1] $F.

 Length of proof is 7.  Level of proof is 6.

 ---------------- PROOF ----------------

 1 [] -P(i(x,y))| -P(x)|P(y).
 2 [] -P(i(i(a,b),i(i(b,c),i(a,c)))).
 3 [] P(i(x,i(y,x))).
 4 [] P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))).
 7 [hyper,1,3,4] P(i(x,i(i(y,i(z,u)),i(i(y,z),i(y,u))))).
 13 [hyper,1,4,7] P(i(i(x,i(y,i(z,u))),i(x,i(i(y,z),i(y,u))))).
 25 [hyper,1,13,13] P(i(i(x,i(y,i(z,u))),i(i(x,i(y,z)),i(x,i(y,u))))).
 29 [hyper,1,13,3] P(i(i(x,y),i(i(z,x),i(z,y)))).
 85 [hyper,1,25,3] P(i(i(i(x,y),i(z,x)),i(i(x,y),i(z,y)))).
 357 [hyper,1,29,85] P(i(i(x,i(i(y,z),i(u,y))),i(x,i(i(y,z),i(u,z))))).
 2044 [hyper,1,357,3] P(i(i(x,y),i(i(y,z),i(x,z)))).
 2045 [binary,2044.1,2.1] $F.

Easy question on Logic and Modes Ponens

1 Answers1