P-adic Valuation and Chebyshev Function Lemma Proof

Question

all following content subject to prompt and drastic changes required due to author stupidity and or lazyness

I am trying to rigorously prove / disprove the following lemma, (as labelled $((0.1),(1.1),(1.2))$)

EDIT: 10/10/2018:

The two primary points of concern remaining are:

1) The complete set of factors for the denominator of the expression inside the fractional part brackets of $(1.2)$ and $(2.2)$

2) Most importantly, if any of what I have stated here is to be true, there must be a true statement regarding the number of prime numbers that exist between $\pi(n)+1$ and $\pi(n+1)$ that exists. I haven't been able to work out how to state this, but the assurance it must exist would be a requisite of any of the piece wise evaluations that assert the primality of $n+1$. It will probably be so elementary it's embarrassing.
$$\lambda_0(n)=\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) -\sum _{i=1}^{\pi(n)} \sum _{j=1}^{\pi ( {\lfloor\frac {n}{i}} \rfloor ) } \Bigl\lfloor\frac{ \ln ( {\frac {n}{i}} ) }{ \ln p_{{j}} } \Bigr\rfloor \ln ( p_{{j}})\tag{0} $$

$${\Biggl\{(2-\delta(n,1))\frac{n!}{\operatorname{e}^{\lambda_0(n)}}}\Biggr\}=0 \Rightarrow\lambda_0 \left( n \right) =\ln \left( \alpha_0 \left( n \right) \right) \land \left\{ {\frac {n!}{\alpha_0 \left( n \right) }} \right\} =0\tag{0.1} $$

$\delta(x,y)$ is the Kronecker delta function.

${\{x}\}$ is the fractional part of $x$.

$\lambda_1(n)$ is the sum of the products of the natural logarithms of the $k^{th}$ prime numbers and the differences between the $p_{k}$-adic valuations of $((n+1)^2)!)!$ and $((n)^2)!)!$ for $k$ from $\pi(n)+1$ to $\pi(n+1)$ inclusively:

$$\lambda_1(n)=\sum _{k=\pi (n ) +1}^{\pi (n+1 ) }\Biggl(\nu_{{p_{k}}}( ((n+1)^2)! ) -\nu_{{p_{k}}} ( n^2!) \Biggr)\ln(p_{k})\tag{1}$$

$$\lambda_1(n)=0 \operatorname{iff} n+1 \not \in \mathbb P\tag{1.1}$$

$${\Biggl\{\frac{\lambda_1(n)}{({\prod^{\pi(n)}_{j=1}}p_j)^2\ln(p_{\pi(n+1)})}}\Biggr\}=0\tag{1.2}$$

$\lambda_2(n)$ is the sum of the products of the natural logarithms of the $k^{th}$ odd prime numbers and the differences between the $p_{k}$-adic valuations of $(n+1)!!$ and $n!!$ over all $p_{k+1}$ for $k$ from $\pi(n)+1$ to $\pi(n+1)$ inclusively:

$$\lambda_2(n)=\sum _{k=\pi (n ) +1}^{\pi (n+1 ) }\Biggl(\nu_{{p_{k}}}( (n+1)! ) -\nu_{{p_{k}}} ( n!) \Biggr)\ln(p_{k})\tag{2}$$

$$\lambda_2(n)=0 \operatorname{iff} n+1 \not \in \mathbb P\tag{2.1}$$

$${\Biggl\{\frac{\lambda_2(n)}{({\prod^{\pi(n)}_{j=1}}p_j)\ln(p_{\pi(n+1)})}}\Biggr\}=0\tag{2.2}$$

$\lambda_3(n)$ is the sum of the products of the natural logarithms of the $k^{th}$ prime numbers $p_{k}$ and the differences between the $p_{k}$-adic valuations of $(n+1)^2!$ and $n^2!$ for $k$ from $\pi(n)+1$ to $\pi(n+1)$ inclusively:

$$\lambda_3(n)=\sum _{k=\pi (n ) +1}^{\pi (n+1 ) }\Biggl(\nu_{{p_{k}}}( (n+1)^2 ) -\nu_{{p_{k}}} ( n^2) \Biggr)\ln(p_{k})\tag{3}$$

$$\lambda_3(n)=\cases{3\ln(p_{\pi(n+1)})&$\,n+1\in \mathbb P $\cr 0& $n+1\not\in \mathbb P $\cr}\tag{3.1}$$

$\lambda_4(n)$ is the sum of the products of the natural logarithms of the $k^{th}$ odd prime numbers $p_{k}$ and the differences between the $p_{k}$-adic valuations of $(n+1)!$ and $n!$ for $k$ from $\pi(n)+1$ to $\pi(n+1)$ inclusively:

$$\lambda_4(n)=\sum _{k=\pi (n ) +1}^{\pi (n+1 ) }\Biggl(\nu_{{p_{k}}}( (n+1) ) -\nu_{{p_{k}}} ( n) \Biggr)\ln(p_{k})\tag{4}$$

$$\lambda_4(n)=\cases{\ln(p_{\pi(n+1)})&$\,n+1\in \mathbb P $\cr 0& $n+1\not\in \mathbb P $\cr}$$ $$\tag{4.1}$$

So far the current draft of the considerations leading to (0.1),(1.1)&(1.2):

Legendre's formula for the p-adic valuation of the factorial of $n$:

$$\nu_{{p}} \left( n \right) =\sum _{j=1}^{ \lfloor {\frac { \ln n }{\ln \left( p \right) }} \rfloor +1} \Bigl\lfloor {\frac {n}{{p}^{j}}} \Bigr\rfloor\tag{i}$$

Natural logarithm Sum to product identity:

$$\ln \left( n! \right) =\sum _{j=1}^{n}\ln \left( j \right)\tag{ii} $$

Unique prime factorization Definition of the factorial of $n$:

$$n!=\prod _{k=1}^{\pi \left( n \right) }{p_{{k}}}^{\nu_{{p_{{k}}}} \left( n! \right) } \tag{iii} $$

An identity for the natural logarithm of the factorial of the nearest integer to $x$ in terms of the first Chebyshev function (used in the first line for the proof of Bertrand's Postulate):

$$\ln ( [x] ! ) =\sum _{k=1}^{\pi ( \lfloor x \rfloor ) }\psi \Bigl( {\frac {x}{k}} \Bigr) \tag{iv} $$

$\pi(x)$ is the number of prime numbers less than or equal to $x$.

${\{x}\}$ is the fractional part of $x$.

Collectively the four above lemmas imply the identity:

$$\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) -\sum _{i=1}^{n-1} \sum _{j=1}^{\pi ( {\lfloor\frac {n}{i}} \rfloor ) } \Bigl\lfloor\frac{ \ln ( {\frac {n}{i}} ) }{ \ln p_{{j}} } \Bigr\rfloor \ln ( p_{{j}} ) =0\quad\quad\quad(\operatorname{i \land ii \land iii\land iv})$$

I then took the following approach to furthering the above conclusion:

Consider the following identities:

$$\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n +1) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n+1}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) -\sum _{i=1}^{\pi(n)} \sum _{j=1}^{\pi ( {\lfloor\frac {n}{i}} \rfloor ) } \Bigl\lfloor\frac{ \ln ( {\frac {n}{i}} ) }{ \ln p_{{j}} } \Bigr\rfloor \ln ( p_{{j}}) =0\tag{v}$$

$$\sum _{i=1}^{\pi (n+1 ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n +1) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n+1}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) -\sum _{i=1}^{\pi(n)} \sum _{j=1}^{\pi ( {\lfloor\frac {n}{i}} \rfloor ) } \Bigl\lfloor\frac{ \ln ( {\frac {n}{i}} ) }{ \ln p_{{j}} } \Bigr\rfloor \ln ( p_{{j}} ) =\ln(n+1)\tag{vi}$$

$$\sum _{i=1}^{\pi (n+1 ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) -\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) =0\tag{vii}$$ $$\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}})=\ln(n!)\tag{viii}$$ $$\sum _{i=1}^{\pi (n+1 ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n+1 ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n+1}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) -\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) =\ln(n+1)\tag{ix}$$

$$\sum _{i=1}^{\pi (n+1 ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}})=\ln(n!)\tag{x} $$

$$\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n+1 ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n+1}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}})=\ln((n+1)!)\tag{xi} $$

$$\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n+1 ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n+1}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}})-\sum _{i=1}^{n } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}})=\sum _{j=1}^{N}\alpha_{g(n,j)}\ln(p _{f(n,j)})\delta\Biggl(\frac{n}{2},\Bigl\lfloor \frac{n}{2}\Bigr\rfloor\Biggr)$$

$${\{g(n,j),f(n,j)}\} \subset \mathbb N\tag{xii}$$

Collectively imply:

$$\sum _{i=\pi (n )+1}^{\pi (n+1 ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n +1) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n+1}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}})=\ln(n+1)\delta(\tau(n+1),2)\quad\quad\quad\quad\quad\quad\quad\quad\tag{0.2} $$

$\tau(k)$ is the total number of divisors of $k$.

$\delta(x,y)$ is the Kronecker delta function.