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 $\set {3, 5, 7}$.

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


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


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

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

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