5

The Deduction Theorem states that for a set of assumptions $\Delta$ and two wffs $A$ and $B$, we have the metalogical relationship:

$$\Delta \cup \{A\} \vdash B \implies \Delta \vdash A \to B$$

In other words if we can prove $B$ from some set of assumptions (conjoined with $A$), then it's the same as proving $A \to B$ from our assumptions.

My question here is not asking for a proof: I'm asking what it's allowing us to do. I'm not even fully sure I understand what this is saying or how its making our lives any easier.

From my uninitiated perspective it would seem that the advantage is letting us treat the logical connective $\to$ as a way to represent a proof from $A$ to $B$ within our logic system rather than mucking about in the metalogical realm, but I feel like I'm looking at it wrong / interpreting this result incorrectly.

What is the deduction theorem telling us? How does it make things more "natural"? What is it doing?

user525966
  • 5,957
  • 1
    In the line below the displayed equation the phrase (intersected with $A$) should read in conjunction with $A$). $\Delta \cup { A } $ means you are adding something to $\Delta$. – Jay Sep 19 '18 at 22:40
  • 1
    It's useful when proving all sorts of things. It's usually a small intermediate step when trying to prove certain things; but it is also used in the proof of the soundness theorem. – Squirtle Sep 19 '18 at 22:40
  • 2
    you just need to go and practice writing proofs using this system without using any additional theorems like this one. Only axioms and MP. You will quickly see that you might want to use some shortcuts like this theorem. – famesyasd Sep 19 '18 at 23:35
  • 1
    After doing enough proofs in the Hilbert system you might want to take a look at another systems to see if they are any more comfortable than this one. – famesyasd Sep 19 '18 at 23:41

1 Answers1

7

The theorem says that to prove an implication it is enough to assume the hypothesis and proceed to prove the conclusion. Proofs of that kind tend to be more natural than proofs that conclude the implication directly.

Just as in regular mathematical practice: many theorems have the form "Assuming $A$, then we have $B$", and we usually prove them by assuming $A$ and using it along the way to conclude $B$. And we feel we are done, even though the task was not to prove $B$, but instead "$A$ implies $B$".

So the theorem is saying that the rules of propositional calculus "capture" a common way of reasoning in practice. The more of such rules we manage to establish the easier it becomes to use the calculus to prove statements. The first few formal proofs always feel strange, perhaps even artificial. These results, such as the deduction theorem, aim at turning this rigid, "artificial" proof system into a natural and useful tool.

There is another key reason why the theorem is useful, now in the context of propositional logic itself, namely, it is a basic tool that helps us prove the completeness theorem. We establish other metatheorems along the way (such as: We can argue by contradiction), and all combined help us in making feasible the task of showing that the proof system is complete: We define a notion of truth for propositional formulas (via truth tables), and completeness says that the proof system is enough to deduce all true implications. This is a very useful first step in more elaborate contexts where we may want to mechanize proofs in more complicated logics.

Andrés E. Caicedo
  • 81,687
  • Wonderful answer, +1! – Taroccoesbrocco Sep 19 '18 at 22:47
  • I am still a little confused as to what this means. If we didn't have the deduction theorem, what would proofs look like instead? "Assuming $A$ we then prove $B$" sounds like the lefthand part of the deduction theorem (before we transform it into the $A \to B$ form), is it not? I'm still unclear what the actual benefit / difference is. – user525966 Sep 19 '18 at 22:50
  • A (formal) proof of $A\to B$ is a chain of deductions from the axioms that ends with $A\to B$. The deduction theorem tells you that you can assume $A$ and use it along the way, and that rather than aiming for $A\to B$ you can focus your energy in actually getting $B$. If you didn't have access to this shortcut, your proof would have to conclude the implication directly. There is nothing wrong with that, and there are some interesting results that you prove by actually aiming for the implication. But it is useful to be able to assume $A$, and more natural in the vast majority of cases. – Andrés E. Caicedo Sep 19 '18 at 23:01
  • Think, for instance, about the proof of $p\to p$. How do you prove that directly? How would the proof go via the deduction theorem? (This is not the best example, by the way, but it may help you see its usefulness. ) – Andrés E. Caicedo Sep 19 '18 at 23:04
  • I'm afraid I still don't understand. For simplicity, assuming $\Delta$ is empty: The left side is saying we assume $A$ and prove $B$, and the right side is saying we prove $A \to B$ from nothing. I don't understand the difference or usefulness. – user525966 Sep 19 '18 at 23:05
  • From axioms and such $ \vdash p \to p$ is proven like so: https://proofwiki.org/wiki/Law_of_Identity/Formulation_2/Proof_3 – user525966 Sep 19 '18 at 23:06
  • 1
    @user525966 As an example, Lemma 1.11(d) in Mendelson's logic text is proven with the assistance of the deduction theorem. An earlier StackExchange question asked for assistance in proving this result without the deduction theorem; the answers given there should be enlightening in understanding the difference. –  Sep 19 '18 at 23:09
  • My eyes are not good enough to read that really thin cursive text, all the letters look the same to me, sadly! I'll look at the linked answer though, thanks – user525966 Sep 19 '18 at 23:13
  • Anyway, there is a technical reason why the theorem is useful, in the context of propositional logic proper: ot is a very useful tool towards the proof that the proof system is complete, that is, it proves all tautologies or, more generally, all true consequences of whatever assumptions we begin with. I'll add this to the answer proper a bit later. – Andrés E. Caicedo Sep 19 '18 at 23:18
  • @user525966 Yep that's the direct proof. And the proof using the deduction theorem is much simpler. We can show ${p}\vdash p$ in one line, and then the deduction theorem gives us $\vdash p\to p.$ – spaceisdarkgreen Sep 19 '18 at 23:18
  • If you use the deduction theorem to prove $A \rightarrow B$, at or near the start you say, "Assume $A$." Then at the end you say "Therefore, $B$." Then you conclude $A \rightarrow B$.

    Without the deduction theorem you don't say "Assume $A$" at all. You start with e.g. "We've already proved $(B \rightarrow (A \rightarrow B)) \rightarrow (A \rightarrow B)$. And we've also proved $(A \rightarrow (B \vee \neg C)) \vee D$..." and so on with other axioms and existing lemmata that you push around until $A \rightarrow B$ pops out. At no point do you "use the assumption" because you never make one.

     –  Sep 19 '18 at 23:25
  • @user525966 Perhaps this is not the best example in some ways, since as we discussed not long ago, most proofs I've seen of the deduction theorem use the fact that $\vdash p\to p.$ (But it still is probably the best way to highlight how deduction theorem proofs are generally easier and more natural than direct proofs of implications.) – spaceisdarkgreen Sep 19 '18 at 23:30
  • @Ian Why so? Doesn't ${A} \vdash B$ literally mean "Assume $A$. Then by the end, we conclude $B$" – user525966 Sep 19 '18 at 23:46
  • 1
    @user525966 Yes, that's what it means. And the deduction theorem says precisely that given a proof for ${A} \vdash B$ ("Assume $A$; by the end we conclude $B$"), there exists also a proof for $\vdash A \rightarrow B$ ("Assume nothing; by the end we conclude $A \rightarrow B$"). –  Sep 20 '18 at 00:21
  • @Ian My point with that question was that without the deduction theorem we're left with ${A} \vdash B$ right? From $A$ we conclude $B$? What's so limiting about that? I guess I just don't see the advantage. What's so helpful about being able to say "from nothing we conclude $A \to B$"? – user525966 Sep 20 '18 at 00:45
  • 1
    No. Without the deduction theorem, you are left with the task of proving $A\to B$. That is what you want. The fact that $A\vdash B$ does not mean a priori that you indeed have proved $A\to B$. Again: Most mathematical theorems in practice are statements of the form $A\to B$. That is what you want to prove. The deduction theorem gives you a way to conclude that, from the sort of move that is standard in mathematics. Perhaps the confusion is that it may seem that the content of the result is that if $\vdash A\to B$ then ${A}\vdash B$. But this is trivial. The deduction theorem is the converse. – Andrés E. Caicedo Sep 20 '18 at 00:55
  • I still don't see the difference. Why do we care about proving $A \to B$ from nothing, as opposed to proving $B$ from $A$? I feel like all these comments so far are just basically restating what the deduction theorem is saying, and I already understand what it's saying as-written but I fail to see the usefulness or utility or what this is actually used for. – user525966 Sep 20 '18 at 01:10
  • Like I understand that the deduction theorem is saying "Assume $A$, we can conclude $B$" can be transformed into "Assume nothing, we can prove $A \to B$", but my question is, "so what?" Why do we care that we can do this? How does this make anything easier? Why do we need this sort of result? How are we using it and how would things be harder if we didn't use it? – user525966 Sep 20 '18 at 01:13
  • 2
    I'm sorry, but it seems to me you are not listening to the answers. Again, in mathematics you prove theorems. You cannot prove assumptions. In proofs, all steps are axioms or theorems. No step is an assumption. Without the deduction theorem you cannot prove "If $X$ is a closed and bounded set of reals, then $X$ is compact" by saying "Assume $X$ is a closed and bounded set of reals" and going from there, because "$X$ is a closed and bounded set of reals" is not itself a theorem. – Andrés E. Caicedo Sep 20 '18 at 01:19
  • @AndrésE.Caicedo I don't follow that -- the deduction theorem proof does mention assumptions being valid possibilities for $\varphi_k$ (in addition to modus ponens conclusions and/or axioms and theorems). And I see things like "Assume $X$ is a bla bla bla" all the time in proofs. It seems like this is more of a semantics claim? "If $A$ then $B$", who's to say that can't be how we interpret $A \vdash B$ in terms of structuring proofs? – user525966 Sep 20 '18 at 01:52
  • Ok. Let's build from there. You are right that in proofs you see assumptions being introduced all the time. The deduction theorem allows you to do that. So, imagine you do not have the deduction theorem. As you say, we can, rather than prove things of the form $\vdash A\to B$, prove the easier things of the form ${A}\vdash B$. This means that the statement "If $X$ is a closed and bounded set of reals, then $X$ is compact" is not a theorem you can use in the middle of proofs. If it happens that you have an assumption "$X$ is a closed and bounded set", then you can conclude "$X$ is compact"... – Andrés E. Caicedo Sep 20 '18 at 02:56
  • ... but if you don't have such an assumption you can't. So now you do not have a theorem that says, say, "if $X$ is compact then every sequence in $X$ has a convergent subsequence", even if you have a theorem that says "if $X$ is closed and bounded, then every sequence in $X$ has a convergent subsequence". The point is that you have effectively lost a lot of maneuvering, and proofs become cumbersome, and probably you cannot recover many of them. You may then decide to change the meaning of formal proof, to allow the introduction of assumptions in the middle of proofs, as need be... – Andrés E. Caicedo Sep 20 '18 at 03:05
  • ... but then you have effectively changed the rules of your proof system, and you can no longer assume the rules you know of the Hilbert-style system you are currently using. And, in truth, it will be very hard to come up with an appropriate proof system for your daily mathematical needs if the system does not have a deduction theorem. You have lost the ability of efficiently handle implications. And implications make up the majority of mathematical statements and ways of reasoning internally in mathematical arguments. – Andrés E. Caicedo Sep 20 '18 at 03:08
  • But it is true that if you are willing to sacrifice a close resemblance between your formal proof system and mathematical reasoning, there are alternatives that you may find more palatable in the way that assumptions are treated or conclusions are drawn--there are many formal proof systems used in practice, Hilbert-style calculi are only one of many alternatives. As I said in the answer, what matters at the end of the day is that your proof calculus is semantically complete. (And, naturally, if you are interested in complexity issues, you may also want your system to be somewhat efficient.) – Andrés E. Caicedo Sep 20 '18 at 03:24
  • "The theorem says that to prove an implication it is enough to assume the hypothesis and proceed to prove the conclusion." No, it does NOT say that. Nor does it imply that. It implies that a proof can get constructed of (A $\rightarrow$ B) if A $\vdash$ B, under "Gamma" also, and that such a construction can happen via the proof of the Deduction Meta-Theorem. – Doug Spoonwood Sep 20 '18 at 18:26
  • " The deduction theorem tells you that you can assume A and use it along the way, and that rather than aiming for A→B you can focus your energy in actually getting B. " No, the Deduction Meta-Theorem does not change the definition of a (formal) proof. – Doug Spoonwood Sep 20 '18 at 18:28
  • @Doug I am aware. I doubt a rather formal presentation of technicalities would be too useful in this case, though. Thank you for the comments. – Andrés E. Caicedo Sep 20 '18 at 18:52