Checksum function verifying even numbers as the sum of two halves

Does the following checksum function make sense?

I am attempting to show that for all even numbers there exists at least two summands that, when normalized to $\frac{1}{2}$, asymptotically sum to 1:

$\lim \limits_{n \to \infty}\frac{n-1}{2[a+ \varphi(a)]}+\frac{n-1}{2[b+\varphi(b)]}\sim \frac{n}{n}=1$

where $n$ are even numbers $\geq 4$, $(a,b)$ are natural numbers where $a+b=n$ and $2 \leq a\leq b$ and $\varphi(n)$ is Euler's totient function

This result holds iff the normalization is performed on prime summands, otherwise the limiting value diverges slightly above 1.

The idea is putting summands through a kind of checksum function that verifies whether or not they are the 'integral' or 'true' sums of the even number. The checksum in this case being the multiplicative identity of $n$, or 1.

Example: $n$ = 10

Input values: (1, 9), (2, 8), (3, 7), (4, 6), (5, 5)

Of the $\frac{n}{2}$ possible summand pairs, only one (5, 5) has the checksum 1 with the next lowest being (3, 7) $\approx$ 1.246. As $n$ tends to infinity so do the checksums tend to 1 for summand pairs that consist only of primes.