Yesterday, I asked about feedback for a proof of the following theorem For all $\phi$, $\phi \in \Gamma^{*}$ if and only if $\Gamma^{*} \vdash \phi$. My main concern was the first part $(\to)$, which I found to be visibly more shaky compared to the second part $(\leftarrow)$. Today, when I raised this issue with my tutor, he seemed to confirm that concern was justified as he argued that the current claim (for part $(\to)$) is insufficient and that one would have to use proof by contradiction to obtain the desired result. To spell it out more clearly, the first line of the argument should read:

Suppose $\phi \in \Gamma^{*}$ and $\Gamma^{*} \not\vdash \phi$.

Now, I'm unsure how to proceed from this, which is why I'm turing to this forum to see whether or not anyone has any suggestions. My instinct tells me that I should proceed by arguing along the lines of using the properties of consistency and the fact that we have either $\Gamma^{*} \cup \{\phi\}$ or $\Gamma^{*} \cup \{\neg \phi\}$ to be consistent. However, I cannot make this reasoning perfectly clear. Help would be very much appreciated.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.