Proving Theorem 2: At Least 2 Games Played

[narledge]

I need help with proving the theorem below

Axiom 1: Each game is played by two distinct teams.
Axiom 2: There are at least four teams.
Axiom 3: There are at least one game played by each team
Axiom 4: Each distinct team plays each of the other teams at most one time

Theorem 2: At minimum there are two games played
According to axiom 2, there are at least four teams and we will call them team A, B, C, D. Since axiom 1 requires that a game is played between two distinct games so team A could playing team C and team B could play D. Axiom 4 states that each team plays each of the other teams no more than once. Each of the four teams are playing at least one game which is required by axiom 3. Therefore, at minimum there are two games played.

 

[Evgeny.Makarov]

Since axiom 1 requires that a game is played between two distinct games so team A could playing team C and team B could play D.

What if A played B or D and the other two teams played among themselves? What if A played a team that is not B, C or D? A proof must consider all cases.

Axiom 4 states that each team plays each of the other teams no more than once.

This does not help advance the proof since this statement only limits the number of played games from above, while the theorem bounds them from below.

Each of the four teams are playing at least one game which is required by axiom 3. Therefore, at minimum there are two games played.

To justifiably say "therefore" you must consider different cases as described above, and in each case conclude that there are at least two games. Otherwise, this seems like a gap in the proof.

 

[narledge]

Thank you for your help, could your offer any additional assistance for how I could write it to make the proof correct?

 

[Evgeny.Makarov]

I would start as you did: According to axiom 2, there are at least four teams and we will call them A, B, C, D. By Axiom 3, A played a game. Let's call its opponent in this game X. Then there is at least one team Y among B, C and D that is not X. By Axiom 3, Y also played a game, and that game must be different from the one between A and X since Y is distinct from both A and X. Thus, we found at least two games.

Note that I assume something that is not, strictly speaking, stated explicitly in the axioms, namely, that games between two different unordered pairs of teams are distinct. In other words, if X and Y played a game and U and V played a game and $\{X, Y\}\ne\{U, V\}$, then these are two different games.