I am working through Chiswell and Hodges and came across this exercise (2.4.5):
Show that {$\phi$} $\vdash $ $\psi $ iff $ \vdash $ $( \phi \rightarrow \psi )$.
For the first part, my line of thought is this:
Assume {$\phi$} $\vdash $ $\psi $. By the Sequent Rule ($\rightarrow$ I) it follows that $\vdash (\phi \rightarrow \psi)$. (We can always add to our set of assumptions, in this case by adding the empty set. Then we apply our Sequent Rule ($\rightarrow$ I).
I am however stuck with the second part of the biconditional:
Assume $\vdash (\phi \rightarrow \psi)$. Then by "Possible sequent rule A", which states that
If the sequent ($\Gamma \vdash \psi$) is a correct sequent, and every statement in $\Gamma$ is also in $\Delta$, then the sequent ($\Delta \vdash \psi$) is also correct.
we get {$\phi, \psi$} $\vdash \phi \rightarrow \psi$. I can then apply the Axiom Rule to get {$\psi, \phi $} $\vdash \psi$.
But how do I get rid of $\psi$ in my set of assumptions?
Another try was to
use the "Possible sequent rule A" to arrive at {$ \phi $} $\vdash \phi \rightarrow \psi$. But then I don't know how to turn the implication into just $\psi$.
I have another idea: I use the "Possible sequent rule A" on $\vdash (\phi \rightarrow \psi)$ to get
{$\phi$} $\vdash (\phi \rightarrow \psi)$. By the axiom rule I get {$\phi$} $\vdash \phi$. I then use Sequent Rule ($\rightarrow E$) to get ($\phi \cup \phi \vdash \psi)$ which is just ($\phi \vdash \psi)$. Pleae tell me I am at least partially correct somewhere.