Formal proof using the deduction theorem

[Yankel]

Hello everyone,

I am trying to find a proof for:

\[\vdash \left ( \sim \alpha \rightarrow \sim \left ( \sim \alpha \right ) \right )\rightarrow \alpha\]

I am using the L inference system, which includes the modus ponens inference rule, and the axioms and statements attached below. That's the only thing I am suppose to use.

In the book from which I took this question from, the third axiom slightly differ from my third axiom. In the book, the 3rd axiom is:

\[\left ( \sim \beta \rightarrow \sim \alpha \right )\rightarrow \left ( \left ( \sim \beta \rightarrow \alpha \right )\rightarrow \beta \right )\]

While my 3rd axiom is:

\[\left ( \sim \beta \rightarrow \sim \alpha \right )\rightarrow \left ( \alpha \rightarrow \beta \right )\]

In the book, this proof uses the 3rd axiom, and I couldn't find a way to replace their 3rd axiom with mine, nor to prove that my axiom leads to theirs, so I can use it instead.

One more thing to bare in mind, which I forgot to mention and is important, I can (and need to) use the deduction theorem.

In the book, they used the deduction theorem, and then the 3rd axiom, one statement ( \[\sim \alpha \rightarrow \sim \alpha\] ) and twice the MP rule.

Thank you in advance.

View attachment 6067View attachment 6068

 

[Valkarie]

Axioms:1. \[\alpha \rightarrow \left ( \beta \rightarrow \alpha \right )\]2. \[\left ( \alpha \rightarrow \left ( \beta \rightarrow \gamma \right )\right ) \rightarrow \left ( \left ( \alpha \rightarrow \beta \right )\rightarrow \left ( \alpha \rightarrow \gamma \right ) \right )\]3. \[\left ( \sim \beta \rightarrow \sim \alpha \right )\rightarrow \left ( \alpha \rightarrow \beta \right )\]Statements:1. \[\sim \alpha \rightarrow \sim \left ( \sim \alpha \right )\]