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Consider the following proof of the principle of explosion using $\lnot \lnot$ elim:

|Assume $p \land \lnot p$

$\quad$|$p$ (from $\land$ elim)

$\quad$|$\lnot p$ (from $\land$ elim)

$\quad$|Assume $\lnot q$

$\quad$$\quad$|$p$ (restatement)

$\quad$|$\lnot q \to p$ (from $\to$ intro)

$\quad$|Assume $\lnot q$

$\quad$$\quad$|$\lnot p$ (restatement)

$\quad$|$\lnot q \to \lnot p$ (from $\to$ intro)

$\quad$|$\lnot \lnot q$ (from $\lnot$ intro)

$\quad$|$q$ (from $\lnot \lnot$ elim)

|$p \land \lnot p \to q$ (from $\to$ intro)

Consider also the proof of LEM using $\lnot \lnot$ elim:

|Assume $\lnot(p \lor \lnot p)$

$\quad$|$\lnot(p \lor \lnot p)$ (restatement)

|$\lnot(p \lor \lnot p) \to \lnot(p \lor \lnot p)$ (from $\to$ intro)

|Assume $\lnot(p \lor \lnot p)$

$\quad$|Assume $p$

$\quad$$\quad$|$\lnot(p \lor \lnot p)$ (restatement)

$\quad$|$p \to \lnot(p \lor \lnot p)$ (from $\to$ intro)

$\quad$|Assume $p$

$\quad$$\quad$|$p \lor \lnot p$ (from $\lor$ intro)

$\quad$|$p \to p \lor \lnot p$ (from $\to$ intro)

$\quad$|$\lnot p$ (from $\lnot$ intro)

$\quad$|$p \lor \lnot p$ (from $\lor$ intro)

|$\lnot(p \lor \lnot p) \to p \lor \lnot p$ (from $\to$ intro)

|$\lnot \lnot (p \lor \lnot p)$ (from $\lnot$ intro)

|$p \lor \lnot p$ (from $\lnot \lnot$ elim)

And a proof of $\lnot \lnot$ elim if we have LEM and the principle of explosion at our disposal:

|$p \lor \lnot p$ (from LEM)

|Assume $\lnot \lnot p$

$\quad$|Assume $p$

$\quad$$\quad$| $p$ (restatement)

$\quad$|Assume $\lnot p$

$\quad$$\quad$|$\lnot p$ (restatement)

$\quad$$\quad$|$\lnot \lnot p \land \lnot p$ (from $\land$ intro)

$\quad$$\quad$|$\bot$ ($\lnot$ elim)

$\quad$$\quad$|$p$ (from $\bot$ elim / the principle of explosion)

$\quad$|$p$ (from $\lor$ elim on LEM and the two previous assumption cases)

|$\lnot \lnot p \to p$ (from $\to$ intro)

My question

Am I reading this right? These rules appear circular and interdependent. If we are granted LEM and the principle of explosion, we can derive $\lnot \lnot$ elimination, but if we're granted $\lnot \lnot$ elimination, we can derive both LEM and the principle of explosion.

Is this correct or is there a way to derive these rules from some other common means?

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    $\begingroup$ The finer details of the answer depend on the specific rules you are allowed to use in this formalism, however, yes, you appear to be reading this right: these rules can be derived from each other. In most formalisms which are set up for this discussion (i.e. classical logic is not "baked in", and it is a formalism for intuitionistic logic which can be made (more) classical by adding new rules), I would expect the principle of explosion to hold regardless, and LEM and $\neg\neg$-elimination to be equivalent. Perhaps in this specific formalism the principle of explosion is not a given. $\endgroup$ – Mees de Vries Nov 19 '18 at 15:23
  • $\begingroup$ @MeesdeVries I believe intuitionistic logic merely disallows LEM and $\lnot \lnot$ elimination but allows the principle of explosion (which is baked into its system without proof) $\endgroup$ – Brandon L Nov 19 '18 at 15:30
  • $\begingroup$ In your proof you have used the principle : $(p \to q) \to ((p \to \lnot q) \to \lnot p)$ that is equivalent (in a system without $\bot$) to $\lnot$-intro : $(p \to \bot) \vdash \lnot p$ $\endgroup$ – Mauro ALLEGRANZA Nov 19 '18 at 15:56
  • $\begingroup$ @MauroALLEGRANZA I agree, but how is this relevant to the question? Principle of explosion is $\bot \to p$ $\endgroup$ – Brandon L Nov 19 '18 at 16:01
  • $\begingroup$ LEM and DN are equivalent and they are both not allowed in Intuitionsitic Logic; but this one use $\bot \to p$. Thus the first two are not equivalent to $\bot \to p$. $\endgroup$ – Mauro ALLEGRANZA Nov 19 '18 at 16:16
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You can have a logical system with double-negation elimination and without explosion.

And you can have logical system where excluded middle and double-negation elimination are provably equivalent without the use of explosion.

The classical version of Neil Tennant's Core Logic (a sort of relevant logic) is a case in point.

Which just goes to show that the rules are not interdependent in quite the way you claim. What's happening is that you are smuggling in other albeit standard assumptions (e.g. about the correct formulation of the or-elimination rule).

Of course, in a standard classical framework, things fit together rather as you say (indeed explosion can be a derived rule).

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  • $\begingroup$ I don't really understand this answer at all. If our system has double negation elimination and no explosion, how is that enforced if I can re-derive explosion with double negation elimination? $\endgroup$ – Brandon L Nov 19 '18 at 15:49
  • $\begingroup$ And how are the rules not interdependent simply due to things like or elimination? As far as I can tell that's an entirely standard rule in virtually all systems, $p \lor q, (p \vdash r), (q \vdash r) \vdash r$. In what way is this "smuggling" non-interdependence? $\endgroup$ – Brandon L Nov 19 '18 at 15:51
  • $\begingroup$ On or-elimination, an alternative rule is from $\alpha \lor \beta$ and a proof from $\alpha$ to one of $\gamma$ or $\bot$ and a proof from $\beta$ to one of $\gamma$ or $\bot$, derive $\bot$ if both subproofs end in $\bot$ or else $\gamma$. This is arguably a more intuitive rule ("if one disjunct leads to absurdity, go with the other!"), and gives us a proof of $\neg\neg p$ from $p$, assuming excluded middle, without explosion. $\endgroup$ – Peter Smith Nov 19 '18 at 16:02
  • $\begingroup$ I don't understand what you mean by that -- can you provide an example using this rule? I'm not sure what "a proof from α to γ or ⊥" means or "if one disjunct leads to absurdity, go with the other!" or how this is nevertheless done without explosion. $\endgroup$ – Brandon L Nov 19 '18 at 16:05
  • $\begingroup$ See logicmatters.net/2017/09/30/… $\endgroup$ – Peter Smith Nov 19 '18 at 16:07

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