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How can I proof that the following formula is a tautology by using Hilbert calculus?
$ \left( \left( A\rightarrow \left( A\wedge \neg A\right) \right) \rightarrow \left( A\rightarrow A \right) \right) $

I know that I have to define axioms and apply the modus ponens. So I studied the examples in this document, however I do not know how to apply this method on my particular formula.

Any help is highly appreciated.

Mauro ALLEGRANZA
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a_12345
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  • It is necessary to say exactly which Hilbert system you mean. For all we know, every tautology is already an axiom of your Hilbert system, and then the formula has a one-line proof. – Carl Mummert Apr 28 '16 at 13:18
  • Thank you for your answer. Actually I do not have any additional information about the Hilbert system I am supposed to use. So maybe this one-line proof is sufficient for my submission :-) – a_12345 Apr 28 '16 at 13:32
  • I doubt it, somehow. But the answer really does depend on the choice of axioms - there are many Hilbert-style systems with different, but equivalent, sets of axioms. – Carl Mummert Apr 28 '16 at 13:57
  • With Ax.1 and Ax.2 you can easily prove $A \to A$. Then use it and modus ponens with a suitable instance of Ax.1: $A \to (B \to A)$. – Mauro ALLEGRANZA Apr 28 '16 at 19:32
  • See the answr to this post for the derivation of $A \to A$. – Mauro ALLEGRANZA Apr 28 '16 at 19:34
  • Thank you both! I understand it now. – a_12345 Apr 29 '16 at 13:14
  • @CarlMummert {CpCqp, CCpCqrCCpqCpr} under hypothetical syllogism and uniform substitution is not equivalent to {CpCqp, CCpqCCpCqrCpr} under substitution and detachment, even though both sets are equivalent under condensed detachment. {CCCpqrCCrpCsp} under substitution and detachment is not equivalent to {CpCqp, CCpCqrCCpqCpr, CCCpqpp} under antecedent commutation and substitution [that is the rule C$\alpha$C$\beta$$\gamma$ $\vdash$ C$\beta$C$\alpha$$\gamma$], but again the axiom sets under condensed detachment are equivalent. – Doug Spoonwood May 03 '16 at 14:16
  • @CarlMummert Oh... also {CCCpqrCCrpCsp} under antecedent commutation and substitution is not equivalent to {CCCpqrCCrpCsp} under substitution and detachment. Many more examples exist like that. – Doug Spoonwood May 03 '16 at 14:21
  • @Doug Spoonwood: Thanks for the examples, but I am a little slow this morning - could you let me know what they are showing? Since you pinged me I think it must be something I wrote, but I can't remember the relationship. – Carl Mummert May 03 '16 at 15:17
  • @CarlMummert I responded to you saying "But the answer really does depend on the choice of axioms - there are many Hilbert-style systems with different, but equivalent, sets of axioms." It's the sets of axioms and the sets of rules together which are equivalent or not equivalent. – Doug Spoonwood May 03 '16 at 16:19

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Note that axiom 1 allows you to get a conditional which has any antecedent you want so long as the consequent is a theorem. For this reason, we can call axiom 1 'recursive variable prefixing.' Axiom 2 we can call 'self-distribution'.

Thus, use recursive variable prefixing and self-distribution to deduce (A→A). Use recursive variable prefixing and (A→A) to deduce (B→(A→A)). Lastly, apply a substitution to obtain your formula.

Doug Spoonwood
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