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$1.$ $\alpha$ $\rightarrow$$(\beta \rightarrow \alpha)$ --- (Ak)

$2.$ $(\alpha \rightarrow (\beta \rightarrow \gamma)) \rightarrow ((\alpha \rightarrow \beta ) \rightarrow (\alpha \rightarrow \gamma))$ --- (AS)

$3.$ $(\lnot \beta \rightarrow \lnot \alpha) \rightarrow ((\lnot \beta \rightarrow \alpha) \rightarrow \beta)$ --- (A$\lnot$)

I'm told that I can use any formula to prove something, but if I know that $\vdash \alpha \rightarrow \alpha $ why I can't use it to prove anything?

For example : To prove $\lnot (\alpha \rightarrow \beta)\vdash \lnot \beta$

$1.$ I can write $\lnot \beta \rightarrow \lnot \beta $, or,known $\lnot \alpha \rightarrow \alpha \vdash \alpha $,I can say $(\lnot \lnot \beta \rightarrow \lnot \beta)\rightarrow \lnot \beta$.

An other example : to prove $\lnot \alpha \rightarrow \alpha \vdash \alpha $

$1.$ I could say $\alpha \rightarrow \alpha$, all the solutions I provided are not correct, but that's what I understood, can you explain what I'm missing?

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    $\begingroup$ In addition to axioms (like 1...3 above), you also need one or more rules of deduction. $\endgroup$ – GEdgar Feb 3 '18 at 12:27
  • $\begingroup$ Maybe for the first example but not for the second one. My problem is that I don't know how to create formula to use axioms or other stuff $\endgroup$ – Bleeeaa Feb 3 '18 at 12:28
  • $\begingroup$ Use Ax.1 : $\beta \to (\alpha \to \beta)$ and contrapose it. $\endgroup$ – Mauro ALLEGRANZA Feb 3 '18 at 12:28
  • $\begingroup$ For the other one, see similar post. $\endgroup$ – Mauro ALLEGRANZA Feb 3 '18 at 12:30
  • $\begingroup$ why the first derivation is not correct? $\endgroup$ – Bleeeaa Feb 3 '18 at 14:31
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The axiom system you are being asked to use is a standard one -- famously used in the classic textbook Elliott Mendelson, Introduction to Mathematical Logic (many editions, there is bound to be one in the library).

Since you are evidently very unclear what the rules of the deduction game are, you badly need to pause and take a slow and careful look at e.g. Mendelson's text.

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  • $\begingroup$ Okay thank you. I have an other question : is $\alpha \rightarrow \lnot(\beta \rightarrow \alpha)$ still axiom Ak? $\endgroup$ – Bleeeaa Feb 3 '18 at 17:41

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