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i'm wondering - how can i use the deduction property to show those two:

1)$\vdash \forall x (p(x) \to q(x)) \to (\exists x p(x) \to \exists x q(x))$

2)$\vdash \forall x (p(x) \to q(x)) \to (\forall x p(x) \to \forall x q(x))$

if there's a better way to prove it (formal proof), i would be interested in learning how to do so.

it seems quite trivial to understand that if $\vdash \forall x (p(x) \to q(x)$ then must exists an x that follows it ($(∃xp(x)→∃xq(x))$ or to put the $\forall$ inside the brackets.

however, i don't know how to prove it and would really appreciate learning how to do so correctly, even though it might seem trivial.

thank you very much for your help

my attempt with 1) after the given answer for 2):

1) $∀x(p(x)→q(x))$ --- assumed

2) $\exists p(x)$ --- assumed

3)$p(c)$

3) $⊢∀x(p(x)→q(x))$ ---

4) $⊢∀x(p(x)→q(x))→(p(c)→q(c))$ --- quantifier axiom, assumption

5) $p(C)→q(C)$ --- assignment

6) q(c) --- from 2) and 5) by Modus Ponens

7) $\exists x p(c)$ ---

8) $\exists x q(c)$ ---

9) $\exists xp(x)→\exists xq(x)$ --- from 2) and 8) by Deduction Theorem

10) $⊢∀x(p(x)→q(x))→(\exists xp(x)→\exists xq(x))$ --- from 1) and 9) by Deduction Theorem

BeginningMath
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    Which system are you using for formal proof? Hilbert system? Which are the rules/axioms that are you allowed to use? – Taroccoesbrocco Jun 15 '18 at 15:57
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    Deduction theorem is definitely not $$((\vdash A) \to (\vdash B)) \to (\vdash (A \to B))$$ If it were you could just instantiate the above with $A$ being unprovable (but not necessarily false) statement to get an unsound result. The deduction theorem correctly is $$(A \vdash B) \to (\vdash A \to B)$$ (sometimes with restrictions on free variables depending on your logic.) – DanielV Jun 15 '18 at 17:59
  • thank you very much for your responses, i just came home: we're using the hilbert series for those kind of proves. if there's another way to prove it that is more efficient, would appreciate learning it as well – BeginningMath Jun 15 '18 at 18:49
  • Huh, I would have thought (2) would have to be an axiom of a Hilbert-type system in order to prove a "generalization theorem" - it would play roughly the same role as $(p \rightarrow q \rightarrow r) \rightarrow (p \rightarrow q) \rightarrow p \rightarrow r$ does in proving the deduction theorem. (And then another axiom $p \rightarrow \forall x , p$ where $x$ is not a free variable in $p$ would play the same role as $p \rightarrow q \rightarrow p$.) – Daniel Schepler Jun 15 '18 at 19:04
  • For (1) you would want to use the existential elimination axiom, which I think would usually look something like: $(\forall x (p(x) \rightarrow q)) \rightarrow (\exists x , p(x)) \rightarrow q$ where $x$ is not a free variable in $q$. At one point, you will want to apply this with $\exists y , q(y)$ in place of $q$. – Daniel Schepler Jun 15 '18 at 19:20
  • @DanielSchepler see my comment under Mauro’s answer. – spaceisdarkgreen Jun 15 '18 at 19:27
  • @DanielSchepler - tried to combine what you explained along with Mauro's explanation. could you check if i did it correctly? – BeginningMath Jun 15 '18 at 19:48
  • You need a rule for steps 7 and 8... – Mauro ALLEGRANZA Jun 15 '18 at 19:50
  • what would be the correct rule/name for those steps? and is the proof correct except those names? – BeginningMath Jun 15 '18 at 19:57
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    @BeginningMath FYI, Hilbert's logic isn't meant to actually be used by humans. It is just a proof-of-concept sort of thing, to show that the concept of logic is feasible. There are far more natural approaches to logic that a human being would use. – DanielV Jun 16 '18 at 00:31
  • @DanielV I tire of the de-humanization of the people who used/use or worked/work with axiomatic systems. Do you really need to imply that they weren't human beings to make your point? Or were you joking? – Doug Spoonwood Jun 18 '18 at 03:42
  • We need to know exactly what axioms and rules of inference before assessing correctness. Does your system allow the use of meta-theorems such as the Deduction Theorem? Also, you probably need to fully parenthesize every statement to satisfy the definition of a formal proof, since every step has to qualify as well-formed according to the definition. – Doug Spoonwood Jun 18 '18 at 03:51
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    @DougSpoonwood I have a tremendous amount of admiration for David Hilbert, what I said was not meant as a derogatory remark. But the fact remains, the logic he created was a starting point meant to demonstrate that logic is in fact possible at a purely symbolic level. Which is an amazing accomplishment. But it wasn't meant to be a high level convenient language; user friendliness was not a design goal. The only criticism intended was towards people who use Hilbert's logic to introduce students to logic, where you have to spend half the time saying "by this we mean that". – DanielV Jun 18 '18 at 04:09
  • @DanielV Axiomatic systems pretty much can always get used in principle. That doesn't hold so much for other systems. Consequently, there may exist some reason to use that sort of system early on in people's logical studies. – Doug Spoonwood Jun 18 '18 at 04:15

2 Answers2

1

Consider 2) :

1) $∀x(p(x) → q(x))$ --- assumed

2) $∀xp(x)$ --- assumed

3) $\vdash ∀x(p(x) → q(x)) \to (p(x) → q(x))$ --- quantifier axiom

4) $p(x) → q(x)$ --- from 1) and 3) by Modus Ponens

5) $\vdash ∀xp(x) \to p(x)$ --- quantifier axiom

6) $p(x)$ --- from 2) and 5) by Modus Ponens

7) $q(x)$ --- from 4) and 6) by Modus Ponens

8) $∀xq(x)$ --- from 7) by Generalization

9) $∀xp(x) \to ∀xq(x)$ --- from 2) and 8) by Deduction Theorem

10) $\vdash ∀x(p(x) → q(x)) \to (∀xp(x) \to ∀xq(x))$ --- from 1) and 9) by Deduction Theorem

For axioms and rules, see E.Mendelson, Introduction to Mathematical Logic, Ch,2 First-Order Logic.

We can prove 1) with Hilbert-calculus using the equivalence :

$¬∀¬ ≡ ∃$.

From $∀x(p(x)→q(x))$ we derive $p(x)→q(x)$ and then $¬q(x)→¬p(x)$.

Then we have to use Generalization and then apply 2) to get $∀x¬q(x) → ∀x¬p(x)$.

Finally, we use again contraposition to get :

$¬∀x¬p(x) → ¬∀x¬q(x)$.

Mauro ALLEGRANZA
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  • (More for OP than Mauro) Note that 2) is axiom Q6 in the Wikipedia list cited. There is a remark after the axiom list that oftentimes instead of axioms Q6 and Q7, there is a generalization rule. This proof (and hopefully OP’s class) are using that alternative system. – spaceisdarkgreen Jun 15 '18 at 19:24
  • @spaceisdarkgreen Hmm... If I thought it was a bit weird to present a Hilbert-type system, but then immediately prove the Deduction Theorem and use it from then on, instead of just presenting natural deduction in the first place -- then I really don't get why you would want to present a hybrid like that between a Hilbert proof system for $\rightarrow$ and a natural deduction type rule for $\forall$... – Daniel Schepler Jun 15 '18 at 19:37
  • thank you very much for your answer. tried to follow your answer to answer 1). could you check if i did it correctly(it's in the original post) – BeginningMath Jun 15 '18 at 19:47
  • @DanielSchepler I pretty much share your bafflement. – spaceisdarkgreen Jun 15 '18 at 21:28
  • could anyone of you show your way to approaching 1) so i can learn the differences between using hilbert and natural deduction? – BeginningMath Jun 15 '18 at 22:09
  • @BeginningMath: Sure; see my answer. Note that there are other styles of natural deduction, but of the many variants I've seen the most practical one is still Fitch-style. I strongly recommend you first learn Fitch-style natural deduction first, because then when you study Hilbert-style systems you will immediately understand the purpose of every single axiom schema there. – user21820 Jun 16 '18 at 07:37
  • @DanielSchepler: Although I personally do not like starting with Hilbert-style systems (see my answer), there are many logic textbooks that do that, and there is a plausible reason for it: It seems that they are interested in going by the conventional approach to first-order logic, namely that there is only one inference rule and many axiom schemas, and every proof is a finite sequence of formulae each of which is an axiom or the result of MP on some preceding formulae. This is convenient for proving meta-theorems about first-order logic, but useless for actual mathematical work. [cont] – user21820 Jun 17 '18 at 08:03
  • [cont] And that same issue also forces the author to state and prove various meta-theorems for the Hilbert-style system, including the deduction theorem and the generalization theorem among others, otherwise it would be really long and messy to prove further results about first-order logic. A side effect is that exercises for students end up being to use a system that includes not just MP but effectively other inference rules, and often after a while nobody wants to use the original Hilbert-style axioms anymore. [cont] – user21820 Jun 17 '18 at 08:09
  • [cont] An alternative to all this is to fully generalize formal system to theorem generator programs or proof verifier programs. The syntactic complexity of natural deduction systems would not make later theorems more difficult to prove, since all that matters is that the system has a proof verifier program. Specifically, we can then set up a framework for manipulating and running basic programs, just as conventionally we need to set up a framework for manipulating first-order formulae. It is not much harder, and the benefits of full generality are obvious. (CC @spaceisdarkgreen) – user21820 Jun 17 '18 at 08:19
  • @user21820 There are plenty of axiom sets where I simply don't understand the purpose of every axiom, other than completeness from the set of axioms, after studying Fitch-style natural deduction. Just ONE example, {CCpqCCqrCpr, CCNppp, CpCNpq} for propositional calculus. The Deduction Theorem is sometimes not proved even though completeness gets proved, e. g. in Lukasiewicz's textbook or systems where there is no deduction theorem (such as equivalential calculus/some three-valued and infinite-valued logics). Rezus has also claimed that Frege doesn't recognize conditional introduction. – Doug Spoonwood Jun 18 '18 at 03:30
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It's good to see your attempt at (1). Here is the proof in Fitch-style natural deduction:

If $\forall x( p(x) \to q(x) )$:

  If $\exists x( p(x) )$:

    Let $c$ be such that $p(c)$.   [$\exists$-elim]

    $p(c) \to q(c)$.   [$\forall$-elim]

    $q(c)$.   [$\to$-elim]

    $\exists x( q(x) )$.   [$\exists$-intro]

  $\exists x( p(x) ) \to \exists x( q(x) )$.   [$\to$-intro]

$\forall x( p(x) \to q(x) ) \to ( \exists x( p(x) ) \to \exists x( q(x) ) )$.   [$\to$-intro]

As you can see, it is supposed to directly reflect natural reasoning. In general, as DanielV said in a comment, Hilbert-style systems cannot be used by humans for practical mathematical work. In contrast, Fitch-style systems like the one I just used can be used in practice even for highly complex theorems and proofs. An additional bonus is that they are readily understood by any professional mathematician even if they have never seen the system before.

user21820
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  • You can see more example proofs in this Fitch-style system here. – user21820 Jun 16 '18 at 07:33
  • Fitch-style systems cannot get (legally) used in some multi-valued logics and infinite-valued logics. Nor in equivalential calculus. If fuzzy mathematics is an extension of some infinite-valued fuzzy logic, since there is no "deduction theorem" for such fuzzy logics, then Fitch-style systems can't get used for fuzzy mathematics. In contrast, axiomatic Frege-Hilbert systems can always get used without such problems arising, as long as axioms exist. – Doug Spoonwood Jun 18 '18 at 03:39
  • I do NOT think that "If", "Let", "be" "such", and "that" exist in the formal language. They pretty much never do. So, this answer is very far from something that leads the user to make such correctly (the OP's original term). This answer conflates formal language with informal language. – Doug Spoonwood Jun 18 '18 at 04:30
  • thank you very much for your comments @DougSpoonwood. those differences are very interesting and important in my point of view. i am planning to read more regarding fitch vs usage of deduction and hilbert. i have to admit that using fitch is easier to prove, but i also wonder if it is acceptable academically. – BeginningMath Jun 18 '18 at 07:36
  • @BeginningMath: It is acceptable academically. Also Doug is wrong about formal systems. Please see this post. – user21820 Jun 18 '18 at 10:59
  • @BeginningMath Note that user21820 did NOT refer to any argument I made and demonstrate in detail how it fails. He just refers to some question about someone personally from a few years back. Please check your definition of proof carefully. I doubt you'll find anything about the use of English words like "let" or "if", and probably something about a sequence of well-formed formulas. That's more than just dropping parentheses (which also isn't allowable solely from the definition of a proof or well-formed formula usually). – Doug Spoonwood Jun 18 '18 at 11:42
  • @DougSpoonwood: Do you or not remember that a moderator once told you to stop harassing me and other users? If you are now willing to learn logic, say so explicitly and I will tell you specifically why you are wrong. If not, talking further will not be productive. – user21820 Jun 18 '18 at 11:53
  • "let", "be", "such" and "that" are probably not part of any formal language that BeginningMath referred to. That concerns what got written above, not any people. It's a simple matter of logic that if a formal proof consists of a sequence of well-formed formulas following from the axioms and rules of inference only, then a formal proof can only include well-formed formulas. I don't know what sort of "formal logic" defines an informal proof as a formal one, but I wouldn't call it such. – Doug Spoonwood Jun 18 '18 at 12:10
  • Let me repeat myself. Are you now willing to learn logic or not? If not, please stop posting wrong comments. I know logic and formal systems very well, and you are simply wrong about general formal systems. End of story. – user21820 Jun 18 '18 at 12:22
  • The original user said: "if there's a better way to prove it (formal proof [emphasis added])..." Every step in a formal proof has to consist of a well-formed formula. Are you now claiming that the string inside of the quotation marks in the following "p(c)→q(c)" is a well-formed formula? Or that "If ∀x(p(x)→q(x))" is a well-formed formula? Or that "Let c be such that p(c)" is a well-formed formula? Because none of those are well-formed formulas. If you don't know that, then you don't know formal logic. – Doug Spoonwood Jun 18 '18 at 16:32
  • What DougSpoonwood wrote is totally wrong. The phrase "formal proof" can mean different things in different formal systems. He obviously does not know basic logic, and yet claims other people do not know logic. He has also harassed many users that post real content about logic. I've had enough. – user21820 Jun 19 '18 at 11:57