MHB What Is the Minimum Number of Meeting Times Needed for Overlapping Committees?

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To ensure that no committee member is scheduled for overlapping meetings, the computer science department needs to determine the minimum number of distinct meeting times required for six committees. Each committee has varying members, leading to potential conflicts in scheduling. The discussion emphasizes the importance of identifying these overlaps to avoid double-booking. Participants are encouraged to share their progress or initial thoughts to facilitate effective assistance. A clear understanding of the member overlaps is crucial for finding a solution.
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The computer science department has 6 committees that meet once a month. How many different meeting times (in a month) must be used to guarantee that no one is scheduled to be at 2 meetings at the same time, if committees and their members are given below.

C1 = {Daim, Viknesh, Lila}, C2 = {Viknesh, Johanes, Philo}, C3 = {Daim, Lila, Philo},
C4 = {Johanes, Lila, Philo}, C5 = {Daim, Lila}, C6 = {Viknesh, Daim, Philo}
 
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Hello johndunlop and welcome to MHB! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?
 

Thread 'Formal derivation of statement from Peano Arithmetic system'

I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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