Prove that $\beta \rightarrow \neg \neg \beta$ is a theorem using standard axioms 1,2,3 and MP I've proven that $\neg \neg \beta \rightarrow \beta$ is a theorem, but I can't figure out a way to do the same for $\beta \rightarrow \neg \neg \beta$. It seems the proof would use Axiom 2 and the deduction theorem (which allows $\beta$ to be an axiom)--but I've endlessy tried values to no avail. Axiom 1: $A \rightarrow ( B \rightarrow A )$. Axiom 2: $( A \rightarrow ( B \rightarrow C ) ) \rightarrow ( ( A \rightarrow B ) \rightarrow (A \rightarrow C) ) $. Axiom 3: $( \neg B \rightarrow \neg A) \rightarrow ( ( \neg B \rightarrow A) \rightarrow B )$. To clarify: A, B, C, $\alpha$, and $\beta$ are propositions (i.e. assigned True or False). $\rightarrow$ and $\neg$ have the standard logical meanings. Note: $\TeX$ification does not work in the IE9 beta.

I'll use the deduction theorem, so I'll assume $\beta$ and need to prove $\
eg\
eg\beta$.

1. $\beta$ (assumption)

2. $\beta\to (\
eg\
eg\
eg\beta\to\beta)$ (axiom 1)

3. $\
eg\
eg\
eg\beta\to\beta$ (modus ponens using 1 and 2)

4. $\
eg\
eg\
eg\beta\to\
eg\beta$ (you have proved that $\
eg\
eg\beta\to\beta$ is a theorem)

5. $(\
eg\
eg\
eg\beta\to\
eg\beta)\to((\
eg\
eg\
eg\beta\to\beta)\to\
eg\
eg\beta)$ (axiom 3)

6. $(\
eg\
eg\
eg\beta\to\beta)\to\
eg\
eg\beta$ (modus ponens using 4 and 5)

7. $\
eg\
eg\beta$ (modus ponens using 3 and 6)