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How to prove suffixing $(p \to q) \to ((q \to r) \to (p \to r))$ from weakening and self-distribution axioms and MP. So, in system with axioms $$A1. p \to (q \to p))$$ $$A2. (p \to (q \to r)) \to ((p \to q) \to (p \to r))$$ and MP figure, how to prove that suffixing statement is a theorem (without Deduction theorem).

Mauro ALLEGRANZA
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Jelena
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  • Thank you, but this formal proof includes Hypothetical Syllogism. I didn't want that. Hence, using your help with this answer I still need the formal proof of Hypothetical Syllogism from A1, A2 and MP. – Jelena Feb 25 '17 at 16:42
  • An hint in the last part of this post : using-deductive-system-modus-ponens. – Mauro ALLEGRANZA Feb 25 '17 at 17:53
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    There is a constructive algorithm for converting proofs using the Deduction Theorem to proofs not using it, requiring only those two axioms and $p \to p$. Your theorem should be easily provable using the DT, so here's a proof strategy: (1) prove $p \to p$; (2) prove your theorem with DT; (3) convert to a proof that doesn't use DT. For the conversion algorithm, see https://en.wikipedia.org/wiki/Deduction_theorem#Conversion_from_proof_using_the_deduction_meta-theorem_to_axiomatic_proof. – Jonas Kölker Aug 18 '17 at 18:49

2 Answers2

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Here is complete answer. Thanks for help.

  1. $(q \to r) \to (p \to (q \to r))$ A1
  2. $(p \to (q \to r)) \to ((p \to q) \to (p \to r))$ A2
  3. $(2) \to ((q \to r)\to (2))$ A1
  4. $(q \to r) \to (2)$ MP 2,3
  5. $(4) \to ((1) \to ((q \to r) \to ((p \to q) \to (p \to r))))$ A2
  6. $(1) \to ((q \to r) \to ((p \to q) \to (p \to r)))$ MP 4,5
  7. $(q \to r) \to ((p \to q) \to (p \to r))$ MP 1,6
  8. $(7) \to (((q \to r) \to (p \to q)) \to ((q \to r) \to (p \to r)))$ A2
  9. $((q \to r) \to (p \to q)) \to ((q \to r) \to (p \to r))$ MP 7,8
  10. $(p \to q) \to ((q \to r) \to (p \to q))$ A1
  11. $(9) \to ((p \to q) \to (9))$ A1
  12. $(p \to q) \to (9)$ MP 9,11
  13. $(12) \to (((p \to q) \to ((q \to r) \to (p \to q))) \to ((p \to q) \to ((q \to r) \to (p \to r))))$ A2
  14. $((p \to q) \to ((q \to r) \to (p \to q))) \to ((p \to q) \to ((q \to r) \to (p \to r)))$ MP 12,13
  15. $(p \to q) \to ((q \to r) \to (p \to r))$ MP 10,14
Jelena
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  • Very good! Yes, that one isn't so bad. The original one will take a bit longer: it should work out to 15 lines total since twice you need to insert 4 more lines in my second proof. – Bram28 Feb 27 '17 at 15:46
  • @Bram28, you are right, 15 lines, thanks a lot, I have just edited the answer. – Jelena Feb 27 '17 at 15:54
  • Yay! If you wrote it all out some of those lines would be pretty long! :) – Bram28 Feb 27 '17 at 16:07
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Let's first prove Hypothetical Syllogism (HS), i.e. that $\{p \rightarrow q , q \rightarrow r \} \vDash p \rightarrow r)$:

  1. $p \rightarrow q$ Premise

  2. $q \rightarrow r$ Premise

  3. $(q \rightarrow r) \rightarrow (p \rightarrow (q \rightarrow r))$ Axiom 1

  4. $p \rightarrow (q \rightarrow r)$ MP 2,3

  5. $(p \rightarrow (q \rightarrow r)) \rightarrow ((p \rightarrow q) \rightarrow (p \rightarrow r))$ Axiom 2

  6. $(p \rightarrow q) \rightarrow (p \rightarrow r)$ MP 4,5

  7. $p \rightarrow r$ MP 1,6

And now that you have HS:

  1. $(q \rightarrow r) \rightarrow (p \rightarrow (q \rightarrow r))$ Axiom 1

  2. $(p \rightarrow (q \rightarrow r)) \rightarrow ((p \rightarrow q) \rightarrow (p \rightarrow r))$ Axiom 2

  3. $(q \rightarrow r) \rightarrow ((p \rightarrow q) \rightarrow (p \rightarrow r))$ HS 1,2

  4. $((q \rightarrow r) \rightarrow ((p \rightarrow q) \rightarrow (p \rightarrow r))) \rightarrow (((q \rightarrow r) \rightarrow (p \rightarrow q)) \rightarrow ((q \rightarrow r) \rightarrow (p \rightarrow r)))$ Axiom 2

  5. $((q \rightarrow r) \rightarrow (p \rightarrow q)) \rightarrow ((q \rightarrow r) \rightarrow (p \rightarrow r))$ MP 3,4

  6. $(p \rightarrow q) \rightarrow ((q \rightarrow r) \rightarrow (p \rightarrow q))$ Axiom 1

  7. $(p \rightarrow q) \rightarrow ((q \rightarrow r) \rightarrow (p \rightarrow r))$ HS 5,6

Bram28
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  • Thank You very much for answering. This is very clear but my question was to prove that the statement is theorem. So we can't use premises, just axioms. So I have to prove $(q \to r) \to ((p \to q) \to (p \to r))$ without any premise. – Jelena Feb 27 '17 at 13:15
  • If we have a proof of $(q \to r) \to ((p \to q) \to (p \to r)) $, it will be the start (new point 1) of the main proof. And also, at the end at the point 7, HS can't be applied in that form. We can only use HS in the form $(q \to r) \to ((p \to q) \to (p \to r))$, as already proved theorem. – Jelena Feb 27 '17 at 13:52
  • @Jelena My first proof shows how to derive a statement of the form $p \rightarrow r$ whenever you have two statements of the form $p \rightarrow q$ and $q \rightarrow r$. So, when in the second proof I refer to HS, you just need to follow those 5 steps (3 though 7) from the first proof to derive the new statement, but where you have to fill in certain statements for $p$, $q$, and $r$. For example, for the first HS you fill in $q \rightarrow r$ for $p$, $p \rightarrow (q \rightarrow r)$ for $q$, and $(p \rightarrow q) \rightarrow (p \rightarrow r)$ for $q$. Let me know if you still need help. – Bram28 Feb 27 '17 at 14:39
  • may I please You just to take a look on my answer bellow. – Jelena Feb 27 '17 at 14:56
  • I completely understand and thank You a lot. I will write whole answer witch will be long. But that was I wanted. Thanks again. – Jelena Feb 27 '17 at 15:10