2

I was asked by a professor a while ago to prove $(p \wedge \neg p)$ implies $q$. Whether through laziness or cleverness, I came up with the following proof:

  • $p \wedge \neg p$ (by assumption).

  • Assume by way of contradiction $\neg q$.

  • $p \wedge \neg p$, therefore $p$ (I forget what this rule is called).

  • Similarly, $p \wedge \neg p$, therefore $\neg p$.

  • We have both $p$ and $\neg p$, a contradiction.

  • Therefore, our assumption, $\neg q$, must have been false, i.e., $q$ is true, the desired result.

Is this right? I keep going either way about it to myself. On the one hand, it definitely feels circular. On the other hand, it seems contain all the correct elements of a contradiction proof.

EDIT: many commenters seem concerned about what I am allowed or not allowed to "use". I am allowed to use whatever you feel mathematically is correct, and not allowed to use things that aren't. The context of the exercise was not a course in logic, or even in mathematics. It was a computer science class in algorithms. I think the professer just wanted to get a feel for our understanding of logic, and also wanted to demonstrate why inconsistant assumptions can be used to prove anything. I don't even remember if the assignment was graded, or if it was what I got. I do remember that when I presented it on the board that it (obviously) wasn't quite what the professor was looking for.

asmeurer
  • 9,774
  • 4
    A formal proof requires us to know what all your axioms and rules of inference are. Without them, it is impossible to tell. – Arturo Magidin Jun 20 '12 at 20:56
  • 2
    Notice that $p \wedge \neg p \implies \neg q$ is also true, so point 2 doesn't contradict anything. What tools are you allowed to use for the proof? – Karolis Juodelė Jun 20 '12 at 20:57
  • Do you know the truth value of $(p\land \neg p)$? Do you know anything about False $\implies$ something? That might help, if you're allowed to use them. – Rick Decker Jun 20 '12 at 20:58
  • Related. – Asaf Karagila Jun 20 '12 at 20:59
  • 6
    If you are "allowed to use whatever you feel is mathematicall correct", then the assignment is nonsense. One cannot produce a formal proof in the absence of a logical framework. If you are allowed to build whatever framework you want, then just assume the proposition you want to show as an axiom and be done with it. For this assignment to make sense, you must have specific axioms and rules of inference at play (a deductive system). There are deductive systems where this implication is not valid, after all. – Arturo Magidin Jun 20 '12 at 21:09
  • @Arturo Not what I "feel" is mathematically correct, what "is" mathematically correct. There is a commonly accepted framework of logic that all mathematicians work in. This is the one that I'm using. Once again, this was not an assignment in a course on logic. – asmeurer Jun 20 '12 at 21:28
  • 4
    @asmeurer: I quoted what you said. And, in the realm of formal logic, there is no such thing as an absolute "is mathematical correct." As I pointed out, there are deductive systems where the implication you want is invalid; and what is mathematically correct in a particular deductive system will depend on what that particular deductive system is. There are many ways of formalizing deductive systems, but most practicing mathematicians don't care what are the axioms and what the conclusions, they just use all. But for this problem to make sense, you do need to care. – Arturo Magidin Jun 20 '12 at 21:33
  • 2
    @asmeurer (cont). Because there are systems where this rule is an axiom (it's built into the system), in which case, the assignment is like asking you prove, using Peano's Axioms, that Peano's Axioms hold (a silly exercise at best). And there are systems where this is a deduction from the system (systems where this assignment would make sense). Easier in some than in others. If you don't specify what you are doing, then this is pure sophistry, just talking yourself into believing what you believe in the first place. – Arturo Magidin Jun 20 '12 at 21:36
  • You have not shown $\neg q$ is false. Instead you have shown that, given $(p\land \neg p)$, the argument leads to a contradiction. So I would not accept this as a proof. – Henry Jun 20 '12 at 21:39
  • 5
    @asmeurer: "that all mathematicians work in" is incorrect, as it turns out. Constructivists, for example, do not accept proofs by contradiction. Be cautious with universal statements, as they are always false. (That last part was a joke.) – Cameron Buie Jun 20 '12 at 21:42
  • @asmeurer In addition to Arturo's excellent comments here, when you write "Not what I "feel" is mathematically correct, what "is" mathematically correct. There is a commonly accepted framework of logic that all mathematicians work in." there exist several problems. 1. There exists no answer which comes as mathematically correct here. 2. The commonly accepted framework, having only two truth values, doesn't prove anything about what qualifies as mathematically correct. There's nothing wrong with consider a system of logic with more than two truth values, one where (p^~p) isn't always false. – Doug Spoonwood Jun 20 '12 at 23:39

5 Answers5

6

Given the tools you have chosen to work with, you have a valid proof of $p\land \neg p\Rightarrow \neg\neg q$. Formally you would need an extra step where you point out that $\neg\neg q$ is the same as $q$, but that's not the real problem here.

The reason why it feels circular is that what you're being asked to prove is one of several possible logical axioms that can be used to justify proof by contradiction in the first place, so the difference between the rule you use and the conclusion is only very slight. It is then natural to wonder whether you're supposed to prove it from more primitive ingredients than proof by contradiction. However, what those particular ingredients might be varies considerably between formalizations of logic.

Basically one can choose any of quite a lot of possible logical foundations to use as axioms, and then prove that all of the other ones follow as logical consequences of the axioms we have chosen. (For pedants in the audience, this is assuming that the axiom sets in question are all for classical first-order logic, blah blah blah). In ordinary mathematical reasoning one tacitly allows all of those standard consequences as single proof steps that need no explanation because the reader is always expected to be able to formalize them in his own favorite logic if he wants to. However, that won't work here, because then there wouldn't be anything for you to prove! So it is necessary to be explicit about which proof steps one allows here, because otherwise your exercise just implodes.

If one happens to choose an axiom set that includes $p\land \neg p\Rightarrow q$ as an axiom, then there's nothing to prove -- a full and complete proof would consist of the words "it's an axiom". But if you choose not to make it an axiom, then what counts as a valid proof depends critically upon what you do consider axioms.

If you do not have a particular formal system you must do your proof in, then I suspect that the expected solution is simply by truth tables -- for each of the possible 4 combinations of whether $p$ and $q$ are true or false, show that $p\land \neg p\Rightarrow q$ evaluates to true.

Then you have proved "$p\land \neg p\Rightarrow q$ is always true" which is just a more convoluted way of saying that you have proved $p\land \neg p\Rightarrow q$ itself.

hmakholm left over Monica
  • 286,031
  • I think the only logical steps I used are $\neg \neg q \Rightarrow q$ and $p \wedge q \Rightarrow p$. Its probably clear by the way that I am not a logician. I have never taken a course on formal logic (or else I probably would have never come up with such a silly proof). – asmeurer Jun 21 '12 at 02:48
  • Oh and I guess also the rule of proof by contradiction, which I would call assume $q \rightarrow$ reducto ad absurdum $\rightarrow \neg q$. The reducto ad absurdum part doesn't seem to me to really rely on any axiom, just the implicit assumption that the logic system is self consistent. – asmeurer Jun 21 '12 at 02:51
  • 1
    Essentially you're saying: "That's not an axiom; it's just how things work". But that does not really make sense, because the word "axiom" is nothing more nor less than a fancy way of saying "that's just how it works". Every proof step is supposed to be justified either by an axiom or by a rule of inference (the distinction between these two categories is somewhat fluid, however), and if you say of some step that it doesn't need any justification, that just means that you're implicitly relying on some axiom or rule that you haven't enunciated. – hmakholm left over Monica Jun 21 '12 at 11:05
  • 1
    Also, "the implicit assumption that the logic system is consistent" can never be used to justify a proof step. In logic, "consistency" is a property of a system that requires that something (i.e., an contradiction) does not have a proof. If you have a choice between allowing a proof step or not, allowing it would make more proofs legal than there would be without it, and so could never make the system closer to being consistent. Conversely, picking any rule or axiom and arbitrarily disallowing it can never make a system inconsistent that wasn't already so. – hmakholm left over Monica Jun 21 '12 at 11:10
  • 1
    Fianlly, you're using a rule saying "if you assume $A$ and using that assumption can prove $B$, then you have proved $A\Rightarrow B$". This rule is usually known as the Deduction Theorem, because in the trend-setting formalizations of logic it is not a primitive proof rule but can be reduced to a combination of other axioms. – hmakholm left over Monica Jun 21 '12 at 11:13
  • "(For pedants in the audience, this is assuming that the axiom sets in question are all for classical first-order logic, blah blah blah)." As long as you haven't mentioned anything explicitly about semantics, the axiom sets could come for something else than classical first-order logic. Bochvar's external system of three-valued logic (see an Introduction to Many-Valued and Fuzzy Logic by Merrie Bergmann p. 82-83) has the same set of tautologies as classical logic. So, given completeness of such a system, all classical logic proofs work for Bochvar's external system also. – Doug Spoonwood Jun 21 '12 at 11:53
2

Some authors will say that the (p∧¬p) consists of a contradiction, while having p as well as ¬p is not a contradiction. One definition of a contradiction comes as a formula F such that for all possible truth values of F's atomic propositions, F has truth value of false. Under this definition, having p as well as ¬p is not a contradiction, and you can only infer a contradiction from having p as well as ¬p if you have a conjunction introduction rule, or something equivalent to such a rule.

Not all mathematicians use a framework where ((p∧¬p)$\implies$q) will hold. You might want to see the applications of relevance logic here. Mitch points out here that ((p∧¬p)$\implies$q) fails for relevance logic.

Though we can fairly easily guess something about the formation rules of your logical system, we know nothing about the semantics of your logical system, nor anything about the rules of inference of your system, nor anything about the axioms of such a system. Since we don't know the rules of inference and don't know the axioms of your system, we don't know what qualifies as a formal proof and what doesn't qualify as a formal proof for your logical system.

If we had a completeness meta-theorem (if a formula qualifies as a tautology, then it qualifies as a theorem), and knew the semantics of your logical system (the truth values that can get meaningfully assigned to propositions), then we could know what we could prove for your logical system, just from calculations with truth values. We just verify that some formula qualifies as a tautology via truth-value calculations, and then we can invoke the respective completeness meta-theorem which implies all tautologies as theorems. This all presupposes the rules of inference and axioms your logical system adequate with respect to the semantics of your system, whatever they might consist of. But, even given just a completeness meta-theorem, since we know nothing about the semantics of your system, and we have nothing which suggests anything about the semantics of your logical system, we don't even know if a given formula comes as provable or not. Not all logical systems have the same tautologies even those that do have a completeness meta-theorem, let alone the same theorems, since the semantics of logical systems does differ significantly.

Your proof comes as not even wrong, since we have no idea of what consists of a proof for your system.

The statement "((p$\land$$\lnot$p)$\implies$ q)" also comes as not even wrong, since we can't even tell if it consists of a tautology, nor can we tell if it qualifies as a theorem for what you want to talk about.

If you presuppose classical logic with its two truth values, and presuppose that your system has a completeness meta-theorem, in an informal context, you really only need to proceed as follows:

  1. (0^$\lnot$0)=0
  2. (1^$\lnot$1)=0
  3. (0$\implies$Q)=1
  4. So, ((P$\land$$\lnot$P)$\implies$Q)=1.
  5. Completeness Meta-Theorem
  6. Therefore, ((P$\land$$\lnot$P)$\implies$Q) is a theorem.

Really, this proof should work for any adequate system of logic where (0$\implies$P)=1, (P$\land$$\lnot$P)=0, which has a completeness meta-theorem also. But, not all logical systems have this going on, and since we have no idea about your rules of inference and your axioms, nor any idea about your semantics, your proof is not correct in a certain sense where "not correct" just means "other than correct" rather than "wrong" necessarily. In truth, your proof is not wrong either, since we have no idea what right and wrong mean in your context. To repeat, your proof is not even wrong.

Community
  • 1
Doug Spoonwood
  • 11,211
0

You make the assumptions that $(p \wedge \neg p)$ is true and that $\neg q$ is true. When you arrive at the contradiction you know that your assumption is wrong but as a whole not which part of it.

Hint 1: It is not $\neg q$.

Hint 2: Your task is not to prove that $q$ is true but rather the validity of the whole form of $(p\wedge\neg p)\to q$.

valid
  • 164
  • I've proven that $q$ is true given $p \wedge \neg p$. This shows the whole form. – asmeurer Jun 20 '12 at 21:08
  • As you state "Therefore our assumption [..] must have been false". Now, your assumption is not $\neg q$ but rather $(p\wedge\neg p)\wedge\neg q$ so you only know that the latter is wrong and nothing about the validity of the former. Instead you know that $(p\wedge\neg p)$ is wrong or $q$ is wrong. – valid Jun 20 '12 at 21:27
  • 5
    @valid: That's not a valid criticism. In formal systems that incorporate the deduction theorem, the assumption part of the deduction theorem is immune from being identified as responsible for a contradiction, because the possibility that the assumption might be false is taken care of by the semantics of the $\Rightarrow$ in the eventual conclusion. Provided that we have something like the deduction theorem and proof by contradiction, the OP does have a valid proof of $(p\land \neg p)\to\neg\neg q$. However, this is not (in all formal systems) the same as proving $(p\land\neg p)\to q$. – hmakholm left over Monica Jun 20 '12 at 22:09
  • @HenningMakholm: "Valid cristicisim", gold :) Of course you are right insofar as the DT does not address the case of the assumption being invalid. Then again, a system incorperating the DT is not precluded from incorporating proof by contradiction as well so the assumption's immunity may still be at jeopardy. Plus, isn't OP's assumption really $p\wedge\neg p\wedge\neg q$, thus under hypothetical provision of DT and proof by contradiction proving the validity of $(p\wedge\neg p)\wedge\neg q\to\neg((p\wedge\neg p)\wedge\neg q)$? If so, OP's proof was not headed in the intially stated direction. – valid Jun 20 '12 at 23:49
  • Suppose you have a three-valued logic where $\lnot$U=T, where "U" indicates the third truth value and "T" truth. Also suppose that (U $\land$ T)=T. Then it follows that (p $\land$ $\lnot$ p) is true, if p has truth value U. Thus, you can assume that (p∧¬p) is true and ¬q is true, and no contradiction can exist, if say q is false, or takes on the third truth value. So, the suggestion of such a proof here doesn't work, since we have no idea about the system in terms of rules of inference, axioms, or in terms of its semantics. – Doug Spoonwood Jun 20 '12 at 23:50
  • @valid: In most systems the rule is just that every assumption must be discharged somehow. Suppose we're in a system with separate primitive rules for DT and proof-by-contradiction. Then if you produce a contradiction, you can choose freely which of your several open assumptions you will discharge by PoC, concluding that this assumption is false. All assumptions you cannot match up with a contradiction have to be discharged by DT, but if only everything eventually does get discharged, the proof is valid. – hmakholm left over Monica Jun 20 '12 at 23:57
  • @DougSpoonwood: From OP's language I (possibly wrongly so) assumed a certain level of knowledge, namely that of a novice to formal logic, and on that based an answer. Also, I'm not too far from that status myself so I was all set. Yes, you are right, currently it is undecided whether OP's proof works. If in your setting I assume behaviour of $\neg U=U$ and $U\to U=U$ I still don't see the initially wanted proof. Then again simply ruling $(p\wedge\neg p)\vdash q$ would end the hassle. – valid Jun 21 '12 at 00:21
  • @valid "Then again simply ruling (p∧¬p)⊢q would end the hassle." Not necessarily. If you have the deduction theorem you have that p|-q implies |-(p->q). So, if you rule (p∧¬p)⊢q, and you have the deduction theorem, then |-((p∧¬p)->q) will have to follow, or you have an inconsistency in the formulation of the system. Ruling such also doesn't work really. Logical systems don't assume such statements in general, and doing such almost surely would get contested by logicians for various reasons. – Doug Spoonwood Jun 21 '12 at 00:30
  • @HenningMakholm "concluding that this assumption is false" Woah, that sounds awfully powerful. Wouldn't the remaining set of assumptions and their implications have to withstand PoC again, possibly releasing from falsehood previously discharged assumptions? Does "if only everything eventually" mean the same as 'not everything at the same time'? If so, I agree, the proof then is valid. – valid Jun 21 '12 at 00:40
  • @DougSpoonwood "then |-((p∧¬p)->q) will have to follow" Exactly. Wouldn't that end the hassle? Basically I was agreeing with you that circumstances are unclear and assumptions easily led to whatever preferred outcome, I mean $U\wedge T=T$ doesn't strike me as a general assumption either. Besides that I declared how I came up with my answer, Occam's razor and such. – valid Jun 21 '12 at 00:56
  • @valid "then |-((p∧¬p)->q) will have to follow" Exactly. Wouldn't that end the hassle?" No, because if the system comes as sound, then we should have |=((p∧¬p)->q)... in other words ((p∧¬p)->q) is a tautology. But, if (U ∧ T)=T and ¬U=T, then ((p∧¬p)->q)==F, when q==F and p==U. So, ((p∧¬p)->q) is not a tautology. We either need a sound system of rules of inference and axioms which has |-((p∧¬p)->q) (we certainly don't know this with respect to the OP), or we need a semantics which yields |=((p∧¬p)->q) along with completeness. We have neither here. – Doug Spoonwood Jun 21 '12 at 02:05
  • @DougSpoonwood I meant ending as in make the implication provable not withstanding the production of an inconsistent or incomplete system along the way. Both these qualities can well be of lesser importance in a didactic context but if underminable should be explicitly so. Yes, in order to reach consensus we should agree on a sound system. No, to prove the implication we don't need a complete and consistent setting but only an accepted method of proof and the required rules and axioms needed for this specific instance. – valid Jun 21 '12 at 03:07
  • @valid: Each assumption needs to be discharged once, and you only discharge one assumption at a time. When you discharge an assumption by PoC, any other assumptions that are still open will need to be discharged later, and discharging them will make the "awfully powerful" conclusion that the assumption was false invisible from the outside. – hmakholm left over Monica Jun 21 '12 at 10:28
0

By definition : $P \Rightarrow Q $ is $\neg P \vee Q$

Then : $$((p \wedge \neg p ) \Rightarrow q) \Leftrightarrow (\neg (p \wedge \neg p)) \vee q \Leftrightarrow (\neg p \vee p \vee q)$$

The disjonction $\neg P \vee P$ is true forall proposition $P$, then the proposition in the right side is true.

Note: Generally we have : if $P$ is false thens $P \Rightarrow Q$ is true forall $Q$.

Mohamed
  • 3,651
  • I'm not asking how to "correctly" prove it. I know how to do that. I'm asking if my "proof" is a correct one. – asmeurer Jun 20 '12 at 21:10
  • I think your proof is correct but the assumption that '$q$ is false ' is without never utility because the implication is realy independant on $q$ since $p \wedge \neg p$ is false all implications beginning with it are trues. – Mohamed Jun 20 '12 at 21:18
  • Not all logical systems have that (P⇒Q) is definable as (¬P∨Q). Suppose that we can compute (P⇒Q) by min(1, 1-P+Q), where 1 indicates truth, 0 falsity, and truth values can take on any value in [0, 1]. Suppose that we can compute (P∨Q) by max(P, Q), and ¬P by 1-P. Then if P=.5, and Q=.5, (P⇒Q) has truth value of 1, while (¬P∨Q) has truth value of .5. Thus, (P⇒Q) does not always come out equal to (¬P∨Q) in terms of truth values, and thus such a definition does not come as permissible, since all definitions like this must have a corresponding logical equivalence. – Doug Spoonwood Jun 20 '12 at 23:58
  • 1
    @ Doug Spoonwood: This is new to me this discussion about systems of logic in mathematics in my above interventions, I mean the sub-system in two truth values. Thank you for these valuable feedback. – Mohamed Jun 21 '12 at 00:42
-1

The answer is: yes, your proof is correct.

Instead of reductio ad absurdum, your can use the principle of explosion. Falsum $\bot$ follows from $p$ and $\neg p$, and ex falso sequitur quodlibet, so one can conclude $q$ directly from $\bot$. This is trivial though, whereas your proof shows the strength of double negation elimination.

Wouter Stekelenburg
  • 1,266
  • 3
    Using the principle of explosion while attempting to prove the principle of explosion itself is not really convincing, methinks. – hmakholm left over Monica Jun 20 '12 at 22:10
  • 1
    "Falsum ⊥ follows from p and ¬p, and ex falso sequitur quodlibet, so one can conclude q directly from ⊥." No, it doesn't. It only follows given either certain rules of inference and axioms, or a certain semantics (usually both). "This is trivial though, whereas your proof shows the strength of double negation elimination." No, the principle of explosion doesn't even come as wrong for logical systems in general, as some have it, and some don't have it. – Doug Spoonwood Jun 21 '12 at 00:00
  • Spoonwood, do you really think he needs to consider logical systems more general then constructive propositional ones? – Wouter Stekelenburg Jun 26 '12 at 18:17
  • I was using the deduction system in http://www.staff.science.uu.nl/~ooste110/syllabi/setsproofs09.pdf, which is used in a university level introductory course in the foundations of mathematics. – Wouter Stekelenburg Jun 26 '12 at 18:53