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This is a follow up question to this answer by Carl Mummert to the question whether every proof with contradiction can also be proved without contradiction. As Carl Mummert pointed out, there are proofs in classical logic which cannot be proved with intuitionistic logic, i.e. which need the law of the excluded middle.

Is there a way or method to show, that a theorem can just be shown with the law of the excluded middle?

Because in minimal logic there is also no principle of explosion: Is it possible to show/prove, that any proof of a theorem needs the principle of explosion?

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Stephan Kulla
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Do you mean theorems which cannot be proven in intuitionistic logic, or classical proofs of theorems that do not have a mere translation to intuitionistic logic? For an example of the latter, there is the classic elementary proof that "there exist $a$,$b$, both irrational, so that $a^b$ is rational". This considers $a=b=\sqrt{2}$, then says $a^b$ is either rational or irrational and splits the problem into cases, without saying which case actually applies. As I recall the relevant case has actually been identified since, so an intuitionistic proof of this result does exist.

An example of the former is relevant to smooth infinitesimal analysis: one cannot show that $x^2=0$ implies $x=0$ under intuitionistic logic, because trichotomy does not hold without the law of excluded middle.

Ian
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  • Thanks for your answer! I updated my question: I want to know which methods there are to show that every proof of a certain theorem needs the law of excluded middle... – Stephan Kulla Jun 12 '14 at 23:16
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    Do you have reference for the claim what $\forall x\in \mathbb R(x<0\lor x=0\lor x>0)$ does not hold intuitionistically? – Git Gud Jun 12 '14 at 23:16
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    http://www.tc.umn.edu/~hellm001/Publications/MathematicalPluralismSIA.pdf Essentially any other primer on smooth infinitesimal analysis will do the job. This isn't a logic text, of course. – Ian Jun 12 '14 at 23:19
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    @Ian Thanks, also found it here. – Git Gud Jun 12 '14 at 23:26
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    @Git Gud: intuitionistic analysis is compatible with the axiom that all functions are continuous. So they can't prove trichotomy, because if they could then they could construct the function that is $1$ on $0$ and $0$ elsewhere, which is discontinuous. – Carl Mummert Jun 13 '14 at 10:46
  • Unfortunately, these arguments with smooth infinitesimal analysis do not apply here. What is called "$\mathbb{R}$" in SIA is not the set of Dedekind reals (or the set of Cauchy reals, for that matter). For the Dedekind reals and for the Cauchy reals you can intuitionistically prove $x^2 = 0 \Rightarrow x = 0$, though not by appealing to trichotomoy, which indeed is not intuitionistically provable. Instead you do it by observing that $a = 0$ is equivalent to $\neg(a # 0)$, where $a # 0$ ("apartness") means $\exists q \in \mathbb{Q}_{>0}. |a| \geq q$, and that $x # 0 \Rightarrow x^2 # 0$. – Ingo Blechschmidt Jun 17 '20 at 20:25
  • In addition to Carl's proof outline, a fun topological way to show that trichotomoy is not provable is to observe that (1) intuitionistically provable statements are "stable in continuous families" and that (2) being zero is not. I tried to write this up for an audience of philosophers of mathematics here, and am available for any questions or comments you might have! – Ingo Blechschmidt Jun 17 '20 at 20:45
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Make the law of the excluded middle ApNp into your only axiom for classical logic, and select suitable rules of transformation and replacement. This gets done in Appendix D of the textbook Elementary Symbolic Logic by Ulrich and Gustason. Consequently, all theorems in classical logic become consequences of the law of the excluded middle.

One could also make the principle of explosion CKpNpq, or CpCNpq, or CNpCpq into one's only axiom if one has suitable rules.

Doug Spoonwood
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  • When you say "all theorems in classical logic become consequences of the law of the excluded middle", do you mean that everything which can be proved classically can also be proved only with the law of excluded middle and some rules of transformation and replacement? – Stephan Kulla Jun 12 '14 at 23:28
  • @tampis In that system, all proofs start with the law of the excluded middle, and thus you can only prove something with the law of the excluded middle and (some to all of) those rules. – Doug Spoonwood Jun 12 '14 at 23:30
  • If I take a theorem proved in the new system with only the law of excluded middle (+ some replacement rules). Why cannot I prove the same theorem with intuitionistic logic? – Stephan Kulla Jun 12 '14 at 23:37
  • @tampis You might be able to prove the same theorem with intuitionistic logic or you might not. It depends on the theorem. Intuitionistic logic doesn't have the law of the excluded middle. – Doug Spoonwood Jun 13 '14 at 01:23