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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?

Bleeeaa
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1 Answers1

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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.

Peter Smith
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  • Okay thank you. I have an other question : is $\alpha \rightarrow \lnot(\beta \rightarrow \alpha)$ still axiom Ak? – Bleeeaa Feb 03 '18 at 17:41