Prove $T_n$ is (2n-4)-regular. I am asked to prove that every triangular graph $T_n$ is a $(2n-4)$-regular graph for $n\ge 3$. I have constructed the following proof by induction, but it feels like it's missing something. Here it is:
For the base case, a quick inspection of the edge set $$E_{T_3} = \{(1,2)(1,3), (1,2)(2,3), (1,3)(2,3)\}$$ reveals that each vertex has exactly $2(3)-4=2$ neighbours, as required. 
Then assume for $v_1\in V$ of $T_n$ there exists a subset $E'_n\in E$ where $$E'_n=\{v_1v_2, v_1v_3,....., v_1v_{2n-4}\}$$ Now consider the case of $v_1\in V$ of $T_{n+1}$. Write $v_1$ as an unordered pair $(x,y)$. Then, for $T_{n+1}$ $$E'_{n+1} = E'_n\cup\{(x,y)(x,n+1), (x,y)(y,n+1)\}$$ $$=\{v_1v_2,v_1v_3,...,v_1v_{2n-4}, v_1v_{(2n-4)+1}, v_1v_{(2n-4)+2}\}$$ Hence, there are $(2n-4)+2=2(n+1)-4$ elements in $E'_{n+1}$, but since $v_1$ was arbitrary, any $v\in V$ of $T_n$ has $2n-4$ neighbours. 
Thus we have shown by induction that $T_n$ is always a $(2n-4)$-regular graph. 
Is this proof complete/sufficiently rigorous and is there anything I could improve?
 A: 
Is this proof complete/sufficiently rigorous?

You are implicitly assuming the edge set of $T_n$ to be a subset of the edge set of $T_{n+1}$. This may be immediate depending on your definition of the triangular graph $T_n$. But I would prefer to define $T_n$ as the line graph of $K_n$, in which case at least something needs to be said to the tune of "Viewing $K_n$ as a subgraph of $K_{n+1}$, $T_n$ becomes a subgraph of $T_{n+1}$, with $n$ additional vertices of the form $\{i,n+1\}$  for all $1 \le i \le n$, and edges joining $\{i,n+1\}$ to $\{i,j\}$ for all $1 \le j \le n$ with $j\ne i$."
(By the way, most people take "triangular graph" to mean "the graph with three vertices and all edges between them"; using "triangular graph" to mean this is less common.)
The above was just something that I think needs more explanation. But more importantly, you have omitted some cases: there are vertices of $T_{n+1}$ that are not vertices of $T_n$, and for such vertices you cannot just say "we take their $2n-6$ old neighbors, and find two more". You must either


*

*Describe all edges out of a vertex of the form $\{i,n+1\}$, and show that there are $2n-4$ of them, or

*Show that $T_{n+1}$ has sufficient symmetry that proving that one vertex has degree $2n-4$ shows the same for all vertices.



I have constructed the following proof by induction, but it feels like it's missing something. 

Well, I've pointed out one thing that it's missing, but maybe that's not what's causing the feeling. Even if we add on to the proof, it might feel like the induction doesn't explain why $T_n$ is $2n-4$-regular.
A more direct proof that avoids induction might give you more understanding by explaining what the $2n-4$ neighbors of an arbitrary vertex are. Given a vertex $\{x,y\}$ of $T_n$, you can say which other vertices are adjacent to this one, and then show that there are exactly $2n-4$ of them.
