I need to find the most simple argument to show that $\vdash_\mathcal{N}((a\rightarrow ((b\rightarrow c)\rightarrow (\lnot d\rightarrow c)))\rightarrow a)\rightarrow a$, where $\mathcal{N}$ has the following axiom schema and modus ponens:
- $\Gamma\vdash_{\mathcal{N}} \beta$, if $\beta\in\Gamma$
- $\Gamma\vdash_{\mathcal{N}}\alpha \rightarrow \big(\beta \rightarrow\alpha \big)$
- $\Gamma\vdash_{\mathcal{N}}\big({\alpha \rightarrow \big(\beta \rightarrow\gamma \big)}\big) \rightarrow \big({\big(\alpha \rightarrow\beta \big) \rightarrow \big(\alpha \rightarrow\gamma \big)} \big)$
- $\Gamma\vdash_{\mathcal{N}}\big( \lnot \beta \rightarrow \lnot \alpha \big) \rightarrow \big(\alpha \rightarrow \beta \big)$
$\Gamma\vdash_{\mathcal{N}} B$, if $B\in\Gamma$. I know I could do this by constructing a truth table, but that's not very "simple". I've also tried to simplify the formula, but that didn't seem to work. Help is much appreciated.
If you can use exploit the fact that $\mathcal{N}$ is complete, then indeed all you need to do is to show that the statement is valid. This can easily be done by a proof by contradiction that shows that any statement of the form $((\phi \rightarrow \psi) \rightarrow \psi) \rightarrow \psi$ is valid:
Suppose $((\phi\rightarrow \psi) \rightarrow \psi)\rightarrow \psi$ is false. Then $(\phi \rightarrow \psi) \rightarrow \phi$ is true, but $\phi$ is false. But with $\phi$ being false, $\phi \rightarrow \psi$ is true, and hence $(\phi \rightarrow \psi) \rightarrow \phi$ is false and hence we have reached a contradiction. Hence, it is impossible for $((\phi \rightarrow \psi) \rightarrow \phi) \rightarrow \phi$ to be false, and hence it is valid.