No Arithmetic Sequence of 4 Primes with Common Difference 2

Definition
There exist no $n \in \Z_{>0}$ such that $n, n+2, n+4, n+6$ are all prime.

Proof
$S$ is a set of $4$ prime numbers of the form $n, n+2, n+4, n+6$.

$S$ must contain as a subset a prime triplet.

From Prime Triplet is Unique, the only one of these is $\left\{{3, 5, 7}\right\}$.

The only sets of the form $\left\{{n, n+2, n+4, n+6}\right\}$ containing $\left\{{3, 5, 7}\right\}$ are:


 * $(1): \quad \left\{{1, 3, 5, 7}\right\}$: as $1$ is by convention not a prime, then this is not $S$.


 * $(2): \quad \left\{{3, 5, 7, 9}\right\}$: as $9 = 3 \times 3$ is not a prime, then this is not $S$.

There are no more possible $\left\{{n, n+2, n+4, n+6}\right\}$ all prime.

Hence, by roof by Contradiction, $S$ does not exist.