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I'm reading the "Introduction to deep inference" notes https://hal.inria.fr/hal-02390267/document and trying to do the Exercise 1.1.4 from there which is about proving the Peirce's law $((A\to B)\to A)\to A$ in Hilbert System H.

The axioms of the system are following:
1. $A\to(B\to A)$.
2. $(A\to(B\to C))\to(A\to B)\to A\to C$.
3. $A\to(A\lor B)$.
4. $B\to(A\lor B)$.
5. $(A\to C)\to (B\to C) \to ((A\lor B) \to C)$.
6. $(A\land B)\to A$.
7. $(A\land B)\to B$.
8. $\texttt{f}\to A$.
9. $\neg\neg A\to A$.
10. $A\to(B\to(A\land B))$.
And the Modus Ponens Rule of Inference.

So, I would like to do this proof using only these axiom schemes and Modus Ponens but encountered some difficulties. I have an idea to use Axioms 1, 2 and 9, but don't actually know how to connect them in a right way.
Since this is a conclusion the Peirce's law should be the instance of Modus Ponens rule at the end. But I don't know for now what should it be.
Will be grateful for any help.

Peter Smith
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A. G
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    I've some doubts about your proof system... There are no rule relating falsum ($\bot$, $\text f$) with negation. – Mauro ALLEGRANZA Apr 23 '20 at 09:02
  • @MauroALLEGRANZA, I've just taken those axioms from the linked notes (pages 6). Also seems that something is missing there. – A. G Apr 23 '20 at 09:04
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    The typical proof in Hilbert-systems is (using the fundamental Deduction Theorem, provable using axioms 1&2) is to assume $(A \to B) \to A$ and both $A$ and $\lnot A$. Having a contradiction, we derive $B$ using rule 8... provided that we have some way to derive $\text f$ from a contradiciton. – Mauro ALLEGRANZA Apr 23 '20 at 09:05
  • @MauroALLEGRANZA, I think I see how it is done in your way, but can we translate that proof to the explained in this notes? The definition there is that A proof in a Hilbert system is a sequence of formulas $A_0, A_1, A_2,..., A_n$, where for each $0 \leq i \leq n$, the formula $A_i$ is either an axiom, or it follows from $A_j$ and $A_k$ via modus ponens. – A. G Apr 23 '20 at 09:14
  • My Answer here might be helpful: https://math.stackexchange.com/questions/3482453/how-to-prove-a-to-b-to-a-to-a-using-lukasiewiczs-axioms-mp-and-deduct/3482471#3482471 ... although this is based on Lukasciewizc system, whose first two axioms are the same, but third is different from H. – Bram28 Apr 23 '20 at 11:06
  • Axiom 8 $\bot \to A$ exploits a contradiction ... but don't you have an axiom that introduces a contradiction.... something like $A \to (\neg A \to \bot)$ .. you have nothing like that? – Bram28 Apr 23 '20 at 11:13
  • @Bram28, in those notes no...there is nothing about the introduction of contradiction, but I see your point. Thanks for the link. – A. G Apr 23 '20 at 14:39
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    Yeah ... I'm suspecting this system as stated might be incomplete ... I can show that Peirce's Law cannot be derived from 1,2,8 and 9 alone... and I don't see how the other axioms can possibly help. – Bram28 Apr 23 '20 at 14:43

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IMO, there is something missing in the description of the system: something linking $\text f$ with the negation sign: $\lnot$.

The simplest way to add it is to define the latter with the former:

$\lnot A := A \to \text f$.

If so, the derivation of Peirce's Law is straightforward, using Deduction Theorem (provable with MP, Ax,1 and Ax.2: see many similar posts on this site):

1) $(A \to B) \to A$ --- premise

2) $\lnot A$ --- assumption [a]

3) $A$ --- Assumption [b]

4) $\text f$ --- from 2) and 3) by MP, using the definition of $\lnot A$

5) $B$ --- from 4) and Ax.8, by MP

6)$A \to B$ --- from 3) and 5) by DT, discharging assumption [b]

7) $A$ --- from 6) and 1) by MP

8) $\text f$ --- from 2) and 7)

9) $\lnot \lnot A$ --- from 2) and 8) by DT and using again the def of $\lnot A$, discharging assumption [a]

10) $A$ --- from 9) Ax.9, by MP

$((A \to B) \to A) \to A$ --- from 1) and 10) by DT

Mauro ALLEGRANZA
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Peirce's Law is not intuitionistically valid. So yes, you will indeed need at some point to appeal to (9), which is the essentially non-intuitionist axiom.

See, perhaps, this earlier answer: Proof of Peirce's Law in Propositional Calculus Double negation elimination is in effect used at the final step.

Peter Smith
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    Note that $\lnot$ is not defined by the text. Mace4 came up with a model to suggest that from 1, 2, 8, and 9 (((A→B)→A)→A ) is not derivable. I don't think the axioms from positive logic would make this derivable. – Doug Spoonwood Apr 23 '20 at 14:50