Is this proof for $\vdash p\vee\neg p$ correct and what other proof would you suggest?

Question

$\;\bullet\;\neg(p\vee\neg p)$ --- assumption
$\;\bullet\;\neg p\wedge\neg\neg p$ --- DM 1 (De Morgan Law)
$\;\bullet\;\neg p$ --- $\wedge$ elim 2
$\;\bullet\;\neg\neg p$ --- $\wedge$ elim 2
$\;\bullet\;\bot$ --- $\bot$ intro 3,4
$\;p\vee\neg p$ --- RAA 1 - 5

actually I'm learning it from internet and not a specific book, can't quite understand what you mean but RAA is a derived rule and standing for Reductio Ad Absurdum and has a separate proof — Pooria, Sep 21 '17 at 16:31
I mean, what rules can you use? Because $\vdash p \lor \neg p$ is just LEM and is a given in many classical systems. — Kenny Lau, Sep 21 '17 at 16:32
Oh I see law of excluded middle, and you can't use DNE because in its proof LEM is used(in proofwiki)! — Pooria, Sep 21 '17 at 16:33
LEM and DNE and RAA are equivalent in the sense that you can prove one from the other, assuming intuitionistic logic, as demonstrated in my link. — Kenny Lau, Sep 21 '17 at 16:34
So please clarify, whether you are seeking a proof of LEM from RAA or from DNE. — Kenny Lau, Sep 21 '17 at 16:34
Note that the rule you used in your question is actually DNE, not RAA. — Kenny Lau, Sep 21 '17 at 16:36
Isn't DNE standing for double negation elimination?! if so how have I used it?! — Pooria, Sep 21 '17 at 16:38
From 1-5 and $\neg$-intro you will obtain $6. \neg \neg (p \lor \neg p)$, and then you can use DNE to obtain $7. p \lor \neg p$. — Kenny Lau, Sep 21 '17 at 16:40
Double negation elimination means $\lnot \lnot p \equiv p$. And so $(\lnot p\land p) \to \bot)$......... — amWhy, Sep 21 '17 at 16:40
RAA is $\neg \varphi \to \psi, \neg \varphi \to \neg \psi \vdash \varphi$ in my nomenclature @amWhy — Kenny Lau, Sep 21 '17 at 16:42
http://imps.mcmaster.ca/courses/CAS-701-02/contributions/nat-deduction.pdf please take a look here for RAA — Pooria, Sep 21 '17 at 16:43
@Pooria Ugh. In that link you posted $\neg A$ is defined as $A \rightarrow \bot$ ... well, you get a perfectly complete proof system that way, but not a very user-friendly. May I recommend you look for some system that has $\neg$ as a basic connective? I think that'll avoid some of your confusions. — Bram28, Sep 22 '17 at 00:51
See also the post prove that $\vdash p \lor \lnot p$ is true using natural deduction — Mauro ALLEGRANZA, Sep 22 '17 at 06:17

score 2 · Accepted Answer · answered Sep 21 '17 at 16:50

Your proof is alright apart from the fact that you used DM1, which although is true in intuitionistic logic, is usually not a given (it is also not given in the natural deduction system you linked).

This is because DM2 is invalid in intuitionistic logic.

So, here's a proof using RAA but not DM1:

  01. ¬[p∨¬p]   assumption
    02. p       assumption
    03. p∨¬p    ∨intro 02
    04. ⊥       contradiction 03 01
  05. ¬p        ¬intro 02-04
  06. p∨¬p      ∨intro 05
  07. ⊥         contradiction 06 01
08. p∨¬p        RAA 01-07

Is this proof for $\vdash p\vee\neg p$ correct and what other proof would you suggest?

1 Answers1