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I want to prove the formula
$$ \vdash\alpha \to ((\alpha \to \beta) \to \beta) $$
using only the Hilbert-style system (without using the Deduction Theorem).

The axioms are:

  • $A_1(\alpha, \beta) = \alpha \to (\beta \to \alpha)$
  • $A_2(\alpha, \beta, \gamma) = (\alpha \to (\beta \to \gamma)) \to ((\alpha \to \beta) \to (\alpha \to \gamma))$
  • $A_3(\alpha, \beta) = ((\lnot \beta \to \lnot \alpha) \to (\alpha \to \beta))$

And Modus Ponens (MP): from $\alpha$ and $\alpha \to \beta$, infer $\beta$. $$ MP\left(\alpha,\left(\alpha\rightarrow\beta\right)\right)=\beta $$

I've had a couple of ideas, but I got stuck in the end. Here's some of what I tried:

$$ A_1\left(q,r\right)=\left(q\rightarrow\left(r\rightarrow q\right)\right)=\varphi_0 $$ $$ A_1\left(\varphi_0,\alpha\right)=\left(\varphi_0\rightarrow\left(\alpha\rightarrow\varphi_0\right)\right) $$ $$ MP\left(1,2\right)=\left(\alpha\rightarrow\varphi_0\right) $$ $$ A_1\left(\varphi_0,\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\right)=\left(\varphi_0\rightarrow\left(\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\rightarrow\varphi_0\right)\right) $$ $$ \left(\alpha\rightarrow\left(\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\rightarrow\varphi_0\right)\right) $$ $$ A_2(\alpha,\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right),\varphi_0)=\left(\left(\alpha\rightarrow\left(\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\rightarrow\varphi_0\right)\right)\rightarrow\left(\left(\alpha\rightarrow\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\right)\rightarrow\left(\alpha\rightarrow\varphi_0\right)\right)\right) $$ $$ MP\left(5,6\right)=\left(\left(\alpha\rightarrow\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\right)\rightarrow\left(\alpha\rightarrow\varphi_0\right)\right) $$ $$ A_1\left(\alpha,\varphi_0\right)=\left(\alpha\rightarrow\left(\varphi_0\rightarrow\alpha\right)\right) $$ $$ A_2\left(\alpha,\varphi_0,\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\right)=\left(\left(\alpha\rightarrow\left(\varphi_0\rightarrow\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\right)\right)\rightarrow\left(\left(\alpha\rightarrow\varphi_0\right)\rightarrow\left(\alpha\rightarrow\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\right)\right)\right) $$ $$ A_1\left(\varphi_0,\left(\alpha\rightarrow\left(\varphi_0\rightarrow\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\right)\right)\right)=\left(\varphi_0\rightarrow\left(\left(\alpha\rightarrow\left(\varphi_0\rightarrow\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\right)\right)\rightarrow\varphi_0\right)\right) $$ $$ \left(\left(\alpha\rightarrow\left(\varphi_0\rightarrow\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\right)\right)\rightarrow\varphi_0\right) $$ $$ A_1\left(\varphi_0,\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\right)=\left(\varphi_0\rightarrow\left(\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\rightarrow\varphi_0\right)\right) $$ $$ \left(\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\rightarrow\varphi_0\right) $$ $$ A_2(\varphi_0,\left(\alpha\rightarrow\beta\right),\beta)=\left(\left(\varphi_0\rightarrow\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\right)\rightarrow\left(\left(\varphi_0\rightarrow\left(\alpha\rightarrow\beta\right)\right)\rightarrow\left(\varphi_0\rightarrow\beta\right)\right)\right) $$ $$ A_1\left(\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right),\varphi_0\right)=\left(\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\rightarrow\left(\varphi_0\rightarrow\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\right)\right) $$ $$ A_2\left(\alpha,\alpha,\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\right)=\left(\left(\alpha\rightarrow\left(\alpha\rightarrow\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\right)\right)\rightarrow\left(\left(\alpha\rightarrow\alpha\right)\rightarrow\left(\alpha\rightarrow\left(\left(\alpha\rightarrow\beta\right)\rightarrow\beta\right)\right)\right)\right) $$

I got lost trying to connect the pieces and actually derive the target formula.

Any help completing this proof using only axioms and MP would be appreciated!

Tankut Beygu
  • 4,412
Kareem Araide
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  • @tankut-beygu thanks, i edited what i wanted to proof to make it clearer, the difference is, we can't assume A here – Kareem Araide May 28 '25 at 13:25
  • You could give some useful context on what you can prove using HIlbert systems, so we can make appropriate suggestions. Do you have a "personal library" of tautologies you already know how to prove? E.g. would you know how to prove transitivity of implication, $(A \rightarrow B) \rightarrow ((B \rightarrow C) \rightarrow (A \rightarrow C))$ using these three axioms? – Z. A. K. May 28 '25 at 13:29
  • @z-a-k i know how to prove: $$ \vdash(α→α) $$ $$ \vdash((β→γ)→((α→β)→(α→γ))) $$ $$ \vdash((α→β)→((β→γ)→(α→γ))) $$ – Kareem Araide May 28 '25 at 14:14
  • One possible hint: Have you seen a proof of $\vdash (\alpha \rightarrow (\beta \rightarrow \gamma)) \rightarrow (\beta \rightarrow (\alpha \rightarrow \gamma))$? If so, then you can specialize that with $\alpha \rightarrow \beta$ in place of $\alpha$, $\alpha$ in place of $\beta$, and $\beta$ in place of $\gamma$, followed by a modus ponens. – Daniel Schepler May 28 '25 at 14:41
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    Otherwise, the deduction theorem doesn't just make an abstract statement, but actually shows you how to convert a Hilbert-style proof of $\Gamma \cup { \alpha } \vdash \beta$ into a Hilbert-style proof of $\Gamma \vdash (\alpha \rightarrow \beta)$. So, you could take the proof given at the previous link, and apply that procedure to get what you want here. – Daniel Schepler May 28 '25 at 14:43
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    I have a post here where I work through a detailed example of how to systematically change a proof that uses premises to a proof without those premises ... effectively doing what the Deduction Theorem says you can do. So, starting with the 1-step proof that $\beta$ follows from $\alpha \to \beta$ and $\alpha$ using MP, you can use the process I go through in that post to convert that proof to a proof that derives $(\alpha \to \beta) \to \beta$ from $\alpha$, and then you can convert that proof to a proof that derives $\alpha \to ((\alpha \to \beta) \to \beta)$ using no premises at all. – Bram28 May 28 '25 at 15:45
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    Here it is: https://math.stackexchange.com/questions/4607540/axiomatic-proof-of-%e2%8a%a2a%e2%86%92b%e2%86%92%c2%acb%e2%86%92%c2%aca-without-using-the-deduction-theorem/4615857#4615857 – Bram28 May 28 '25 at 15:45

1 Answers1

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This is a derivation without using any common theorems as schema:

  1. $(\alpha\rightarrow\beta)\rightarrow((\beta\rightarrow(\alpha\rightarrow\beta))\rightarrow(\alpha\rightarrow\beta))\tag{Ax1}$
  2. $((\alpha\rightarrow\beta)\rightarrow((\beta\rightarrow(\alpha\rightarrow\beta))\rightarrow(\alpha\rightarrow\beta)))\rightarrow(((\alpha\rightarrow\beta)\rightarrow(\beta\rightarrow(\alpha\rightarrow\beta)))\rightarrow((\alpha\rightarrow\beta)\rightarrow(\alpha\rightarrow\beta)))\tag{Ax2}$
  3. $((\alpha\rightarrow\beta)\rightarrow(\beta\rightarrow(\alpha\rightarrow\beta)))\rightarrow((\alpha\rightarrow\beta)\rightarrow(\alpha\rightarrow\beta))\tag{MP 1, 2}$
  4. $(\alpha\rightarrow\beta)\rightarrow(\beta\rightarrow(\alpha\rightarrow\beta))\tag{Ax1}$
  5. $(\alpha\rightarrow\beta)\rightarrow(\alpha\rightarrow\beta)\tag{MP 3, 4}$
  6. $((\alpha\rightarrow\beta)\rightarrow(\alpha\rightarrow\beta))\rightarrow(((\alpha\rightarrow\beta)\rightarrow\alpha)\rightarrow((\alpha\rightarrow\beta)\rightarrow\beta))\tag{Ax2}$
  7. $((\alpha\rightarrow\beta)\rightarrow\alpha)\rightarrow((\alpha\rightarrow\beta)\rightarrow\beta)\tag{MP 5, 6}$
  8. $\alpha\rightarrow((\alpha\rightarrow\beta)\rightarrow\alpha)\tag{Ax1}$
  9. $(((\alpha\rightarrow\beta)\rightarrow\alpha)\rightarrow((\alpha\rightarrow\beta)\rightarrow\beta))\rightarrow(\alpha\rightarrow(((\alpha\rightarrow\beta)\rightarrow\alpha)\rightarrow((\alpha\rightarrow\beta)\rightarrow\beta)))\tag{Ax1}$
  10. $\alpha\rightarrow(((\alpha\rightarrow\beta)\rightarrow\alpha)\rightarrow((\alpha\rightarrow\beta)\rightarrow\beta))\tag{MP 7,9}$
  11. $(\alpha\rightarrow((\alpha\rightarrow\beta)\rightarrow\alpha)\rightarrow((\alpha\rightarrow\beta)\rightarrow\beta))\rightarrow((\alpha\rightarrow((\alpha\rightarrow\beta)\rightarrow\alpha))\rightarrow(\alpha\rightarrow((\alpha\rightarrow\beta)\rightarrow\beta)))\tag{Ax2}$
  12. $((\alpha\rightarrow((\alpha\rightarrow\beta)\rightarrow\alpha))\rightarrow(\alpha\rightarrow((\alpha\rightarrow\beta)\rightarrow\beta)))\tag{MP 10,11}$
  13. $\alpha\rightarrow((\alpha\rightarrow\beta)\rightarrow\beta)\tag{MP 8, 12}$

Notice that the lines 1-5 constitute the derivation $A\rightarrow A$, and the lines 7-13 constitute the derivation and application of transitivity of implication, that is, $A\rightarrow B, B\rightarrow C\vdash A\rightarrow C$.

Tankut Beygu
  • 4,412