I am currently reading through an 'Intro to the Discharging Method' https://arxiv.org/pdf/1306.4434v1.pdf and am confused about a step in the proof of Lemma 5.3. The lemma itself was originally from this paper if it helps http://link.springer.com/article/10.1007/BF01240258.
A t-alternating cycle alternates between t-vertices and vertices of higher degree.
Lemma 5.3
If G is a simple plane gragh with $\delta (G)\geq 2$ (minimum degree), then G contains
1) An edge uv such that $d(u)+d(v)\leq 15$ or
2) A 2-alternating cycle
The proof starts off by saying, given a counterexample G, all edges must be so that $d(u)+d(v)\geq 16$ therefore both neighbours of any 2-vertex must have degree at least 14. However, it then goes on to say 'Since G is simple' every two vertex lies in a $4^+$-face (that is a face with length at least 4).
I do not understand why a 2-vertex can't lie in a triangular face, I feel like I could be missing something very simple so any pointers would be welcome.
I realised my mistake after asking one of the authors.
Since G is simple, of the TWO faces incident to the edges joining a 2 vertex to two 14+ vertices, at least one of the two faces must be a triangular face.
I was mistaking the statement to be about either of the two faces.