# Maximally Consistent Set (Proof by Contradiction)

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.