# Attached straightforward cubic chart

I am attempting to recognize this trouble and also of course this is from my assignment and also I need to be doing it myself, yet I have actually been looking at it for 2 hrs and also not obtaining anywhere, so determined to upload it below.

Allow G be a linked cubic straightforward chart which contains 2 edge-disjoint extending trees show that|G|= 4.

Let $G$ be a straightforward cubic chart, with 2 side - disjoint extending trees, that has $n$ vertices and also $m$ sides. Given that $G$ is normal of level 3, we have

$3n = 2m.$

Now any kind of extending tree has $n-1$ sides, and also $G$ has 2 disjoint extending trees. So we have

$m \ge 2n - 2.$

Combining the first formula and also the 2nd inequality, we get

$n \le 4$.

Currently there are no cubic straightforward charts on 1, 2 or 3 vertices, so the outcome adheres to. (It is open to question whether there is one on 0 vertices.)

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