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I'm looking to prove that $(P \leftrightarrow \neg P)$ is a contradiction using a natural deduction proof (which is to say, I want a proof to show $(P\leftrightarrow \neg P)\vdash Q$). In case it helps, the specific system I'm working in is as outlined in Halbach's Logic Manual (tree structure, introduction and elimination rules for each connective; see the link below), but it's the overall structure of the proof I'm struggling with.

Given a proof that shows $\vdash (P \vee \neg P)$ I can transform this into the desired proof, but that generates a very large tree given the simplicity of the sentence, because the proof for $\vdash (P \vee \neg P)$ is fairly long itself.

I can't shake the feeling there must be a more straightforward (even if still indirect) proof, but haven't been able to find it so far.

Edit: As found by lemontree, the ruleset I am using is listed here.

dbmag9
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    Would you mind recalling the facts about negation ? – FiMePr Jun 29 '20 at 15:50
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    It would be useful to link to the rule set you're using, if it exists on the web. Is this it? http://users.ox.ac.uk/~logicman/jsslides/ll6p.pdf – Natalie Clarius Jun 29 '20 at 16:05
  • @lemontree That's the set; I'll edit that link into the question. – dbmag9 Jun 29 '20 at 17:23
  • @FiMePr The rules for negation are on the 4th page of the PDF now linked. The introduction rule is: Given $\Gamma, \phi \vdash \psi$ and $\Gamma, \phi \vdash \neg \psi$, you can form a proof that $\Gamma \vdash \neg \phi$ (where $\Gamma$ is a set of sentences and $\phi, \psi$ are sentences). The elimination rule just switches all the $\phi$s for $\neg\phi$s and vice versa. – dbmag9 Jun 29 '20 at 17:28

3 Answers3

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One possible route is proving $\lnot(P \leftrightarrow \lnot P)$ without premises. Following that path, this would be a Gentzen-style proof, I think.

$ \def\be\qquad\mathbf{\leftrightarrow Elim} \def\bi\qquad\mathbf{\leftrightarrow Intro} \def\ne\qquad\mathbf{\neg Elim} \def\ni\qquad\mathbf{\neg Intro} $

$ \begin{equation} \dfrac{ \dfrac{ [P] \qquad [P \leftrightarrow \lnot P] }{ \dfrac{ [P] \qquad \lnot P }{\lnot P}\ni}\be \quad \dfrac{ [\lnot P] \qquad [P \leftrightarrow \lnot P] }{ \dfrac{ [\lnot P] \qquad P }{P}\ne}\be }{\lnot(P \leftrightarrow \lnot P)}\ne \end{equation} $

EDIT:

Based on OP comments, I am adding a proof of $P \leftrightarrow \lnot P \vdash Q\\$.

$ \begin{equation} \dfrac{ \dfrac{ [P] \qquad P \leftrightarrow \lnot P }{ \dfrac{ [P] \qquad \lnot P }{\lnot P}\ni}\be \quad \dfrac{ [\lnot P] \qquad P \leftrightarrow \lnot P }{ \dfrac{ [\lnot P] \qquad P }{P}\ne}\be }{Q}\ne \end{equation} $

F. Zer
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    Thanks for this; in fact your tree can be adapted into a tree for $(P\leftrightarrow \neg P) \vdash Q$ just by taking the square brackets off the occurences of $(P\leftrightarrow \neg P)$ in the top row, and replacing the final sentence with $Q$ (using $\neg$Elim to 'discharge' an assumption of $\neg Q$ that does not feature in the tree). – dbmag9 Jun 30 '20 at 13:16
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    You're welcome. Good observation. I'll add it to the answer. – F. Zer Jun 30 '20 at 13:46
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You can prove Lemma 1: $P\to\lnot P\vdash\lnot P$:

  1. $P\to\lnot P$ - assumption
  2. | $P$ - additional assumption
  3. | $\lnot P$ - modus ponens on 1. and 2.
  4. | $\bot$ - elimination of negation from 2. and 3.
  5. $\lnot P$ - introduction of negation on 2-4.

and similarly Lemma 2: $\lnot P\to P\vdash P$. Now:

  1. $P\leftrightarrow\lnot P$ - assumption
  2. $P\to\lnot P$ - elimination of equivalence from 1.
  3. $\lnot P\to P$ - elimination of equivalence from 1.
  4. $P$ - Lemma 2 and 3.
  5. $\lnot P$ - Lemma 1 and 2.
  6. $\bot$ - elimination of negation from 5. and 6.
  7. $Q$ - ex falso quodlibet from 6.
SMM
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    The rules for $\leftrightarrow$ in the source specified by the OP are defined such that from 1. and 4. one obtains 5. directly without the intermediate step 2. (and analogous for the other direction). – Natalie Clarius Jun 29 '20 at 16:07
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(Posted after answer accepted.) Using a form of natural deduction:

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Dan Christensen
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