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Show that $((\phi → \psi)→((\psi→\chi)→(\phi→\chi)))$ is a Theorem of L.

I a previous part of the Q i am asked to state the deduction theorem so I assume i have to use this and the axioms A1, A2, A3, and also Modus Ponens to prove that the formula is a theorem of L.

I am really struggling with doing any question using the axioms to show something is a theorem of L. I can almost work my way through an example, but even then I am confused with why / how certain steps are done.

Could you help me work through this. Thanks

hmakholm left over Monica
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ZZS14
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    It would be a lot easier if you reveal to us what L, A1, A2, and A3 are. These names are not standard, but must be specific to the particular text you're working from. – hmakholm left over Monica Apr 22 '14 at 12:36

2 Answers2

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If you are allowed to appeal to the deduction theorem then this is very easy, and indeed you don't need to make further appeal to any of the axioms (so it just happens that this question can be answered even though you haven't told us which particular axioms A1, A2, A3 in fact are).

For you can show

$$\varphi, (\varphi \to \psi), (\psi \to \chi) \vdash_L \chi$$

using modus ponens twice (OK?). Then one application of the deduction theorem gives you

$$(\varphi \to \psi), (\psi \to \chi) \vdash_L (\varphi \to \chi)$$

(OK?) and a second application gives you

$$(\varphi \to \psi) \vdash_L ((\psi \to \chi) \to (\varphi \to \chi))$$

(OK?). And then what happens if you use the deduction theorem again?

Peter Smith
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  • I sort of can see where the deduction theorem has been used, but can you explain why you have started with what you have and what that exactly is, i don't understand the commas.. – ZZS14 Apr 22 '14 at 12:34
  • That's just a list of three premises, separated by commas. But your even asking that suggests you need to do some serious homework, with a textbook or three in the library! – Peter Smith Apr 22 '14 at 12:46
  • i do, i really cannot get my head around this sort of question. I'm finding text books no help. – ZZS14 Apr 22 '14 at 12:52
  • could you explain the steps? – ZZS14 Apr 22 '14 at 13:01
  • @ZZS14 - as per answer of Peter Smith, in this case you do not need axioms (A1)-(A3) [I suspect they are those of Elliott Mendelson, Introduction to mathematical logic (4ed - 1997)] but of course you have used them (more precisely : (A1) and (A2) to prove the Deduction Theorem). You need aslo to "master" the concept pf deduction form assumptions : $\Gamma \vdash \varphi$ : you apply it in the DT to get from $\Gamma \cup { A } \vdash B$ to $\Gamma \vdash $A \rightarrow B$. 1/2 – Mauro ALLEGRANZA Apr 22 '14 at 13:29
  • @ZZS14 - having said that, you proceed as follows; assume the set $\Gamma$ of assumptions : $φ,(φ→ψ),(ψ→χ)$; from $φ$ and $(φ→ψ)$, apply modus ponens and get $ψ$; from $ψ$ and $(ψ→χ)$, apply mp again, to get $χ$. Now uou have a deduction of $χ$ form assumptions $φ,(φ→ψ),(ψ→χ)$, i.e. $φ,(φ→ψ),(ψ→χ) \vdash χ$. Apply DT once and get : $(φ→ψ),(ψ→χ) \vdash (φ→χ)$; apply DT to get $(φ→ψ) \vdash ((ψ→χ)→(φ→χ))$. Finally, apply DT a third time and you will have : $\vdash (φ→ψ) → ((ψ→χ)→(φ→χ))$. – Mauro ALLEGRANZA Apr 22 '14 at 13:34
  • i understand once you use the deduction theorem but i am confused by what you start with and why – ZZS14 Apr 22 '14 at 16:44
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    @ZZS14 - but the starting point is : what is your textbook ? What have you already studied ? If you know the definition of deduction from assumptions, then our set $\Gamma$ of assumption is the set ${ φ,(φ→ψ),(ψ→χ) }$. Form this set we derive $χ$ [see PS's answer]; thus, we have ${ φ,(φ→ψ),(ψ→χ) } \vdash χ$. Now we apply DT : from $\Gamma \cup { A } ⊢ B$ to $\Gamma \vdash A \rightarrow B$, with $φ$ as $A$ and $χ$ as $B$ to get : ${ (φ→ψ),(ψ→χ) } \vdash φ \rightarrow χ$. And so on ... – Mauro ALLEGRANZA Apr 22 '14 at 19:03
  • I've studied a logic course, but it wasn't very well lectured and now the exam in 1st May. I know the definition of what is meant by saying there is a deduction of $\psi$ from $\gamma$? I just don't understand the first step before the use of the deduction theorem – ZZS14 Apr 23 '14 at 07:00
  • @ZZS14 - Please, review Peter Smith's answer: he starts from the set ${ φ,(φ→ψ),(ψ→χ) }$ of assumptions and pick up the first two : $φ$ and $(φ→ψ)$ in order to apply mp; in this way, he derives $ψ$. Then he uses it with the third assumption : $(ψ→χ)$; again, by mp he derives $χ$ , and it's done. – Mauro ALLEGRANZA Apr 23 '14 at 07:46
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So, let's look at "((ϕ→ψ)→((ψ→χ)→(ϕ→χ)))". It is a conditional. What is it's antecedent? (ϕ→ψ). What is it's consequent? ((ψ→χ)→(ϕ→χ)). Thus, if we have the deduction (meta) theorem, if we assume (ϕ→ψ) and can derive ((ψ→χ)→(ϕ→χ)), we can then use the proof procedure outlined by the proof of the deduction theorem to get to ((ϕ→ψ)→((ψ→χ)→(ϕ→χ))). But, how do we get ((ψ→χ)→(ϕ→χ))? Well, what is the primary connective for ((ψ→χ)→(ϕ→χ))? A conditional. So, what's the antecedent? (ψ→χ). What's the consequent? (ϕ→χ). Thus, if we can assume (ψ→χ) and derive (ϕ→χ), then we can infer ((ψ→χ)→(ϕ→χ)) by the derivable rule of conditional introduction, which follows from the deduction theorem. Note that when trying to prove ((ψ→χ)→(ϕ→χ)), we still had the assumption (ϕ→ψ) in place. How do we derive (ϕ→χ)? Well, could we assume the antecedent ϕ and deduce the consequent χ? In other words... can we do this...

  1 |   (ϕ→ψ) assumption
  2 ||  (ψ→χ) assumption
  3 ||| ϕ     assumption
  .
  .
  .
  x ||| χ     ?
Doug Spoonwood
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  • This looks dangerously close to confusing the Deduction Theorem, a meta-level theorem about a Hilbert-style proofs system, with Conditional Proof as a rule inside a natural deduction system allowing the making and discharge of temporary assumptions. – Peter Smith Apr 22 '14 at 13:33
  • @PeterSmith If we have the deduction theorem, then conditional proof is a derivable rule of inference for a Hilbert system. – Doug Spoonwood Apr 22 '14 at 13:41
  • It is rather more delicate an issue than that. What is true is that you can conservatively extended a Hilbert system into a mixed system which allow the making and discharge of temporary assumptions. But that's to introduce a proof-system of a new character, and we are no longer inside e.g. The Mendelson system I guess the OP is using. – Peter Smith Apr 22 '14 at 16:04
  • @PeterSmith How is it a more delicate issue? How is it a proof system of a new character? The relevant subsystem under detachment probably is either {CpCqp, CCpCqrCCpqCpr} or {CpCqp, CCpqCCpCqrCpr} both of which have the same theorems [both axiom sets imply the deduction meta theorem, and if you have the deduction meta theorem, then you have the law of commutation which makes the axiom sets interderivable). The extension here doesn't introduce any new theorems nor remove any theorems. So, how have we stepped outside of the original system? – Doug Spoonwood Apr 22 '14 at 17:30
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    The mixed system allows the temporary introduction of new assumptions; a Frege-Hilbert system doesn't, by definition. – Peter Smith Apr 22 '14 at 18:10