Superposition of two dependent Poisson processes $N_1(t)$ and $N_2(t)$ are two independent Poisson processes with intensities $λ_1$ and $λ_2$ respectively. $v_1$ and $v_2$ are two dependent positive random variables, and $p=P(v_1<c)=P(v_2<c)$, where $c$ is a constant. 
When some random event of $N_1(t)$ happened and $v1<c$, it means an event of point process $N_3(t)$ happened. When some random event happened in $N_2(t)$ and $v_2<c$, it means an event of point process $N_4(t)$ happened.
Then $N_3(t)$ and $N_4(t)$ are two dependent Poisson processes with intensities $p*λ_1$ and $p*λ_2$ respectively. My questions are: 


*

*Is the superposition of $N_3(t)$ and $N_4(t)$ a Poisson process approximately? 

*Let $N_5(t)$ be a Poisson process with intensity $p(λ_1+λ_2)$, is there 
$P\{N_3(t)+N_4(t)\le m\} \le P\{N_5(t)\le m\}$? (Here $m$ is a integer.)


All your answers are helpful. I really appreciate them and thanks for gung's editing. 
Specially, $v_1$ and $v_2$ be denoted as two dependent log-normal varialbles with correlation $\rho$. In other words, we can describe $v_1$ and $v_2$ as a multivariate log-normal variable. In this time, how to answer those two questions?
I agree that ($v_1$, $v_2$) can be replaced by the dependent Bernoulli pair($b_1$, $b_2$). And now $N_{3}(t)$ and $N_{4}(t)$ are two dependent poisson process. Is their superposition a poisson process approximately?
I really appreciate all your reply. 
I have read sveral literature and consulted a professor. $N_{3}(t)$+$N_{4}(t)$ is a Cox process according to reply of the professor. And it converges to a poisson process.(Serfozo 1984, Chen & Xia 2011).
 A: Too long for a comment.

But I do not think v1 and v2 can be replaced with a Bernoulli random variable. Specially, v1 and v2 be denoted as two dependent log-normal varialbles with correlation ρ. In other words, we can describe v1 and v2 as a multivariate log-normal variable. In this time, how to answer those two questions?

I think you're mistaken; they can be. 
The only way that $v_1$ and $v_2$ affect anything is through the condition that they're less than $c$. Let $b_1$ and $b_2$ take the value 1 when their corresponding $v_i<c$. You now have that $b_1$ and $b_2$ are dependent Bernoullis. What information relevant to the question other than the information in the pair $(b_1,b_2)$ is there in $(v_1,v_2)$? If there is no additional information of relevance to the question, then $(b_1,b_2)$ clearly can replace $(v_1,v_2)$.
A: Your question is not well stated
As I claim in my comments, according to your description of the problem, $N_3$ and $N_4$ are not Poisson processes. Indeed (according to your description), 
$$N_3(t) = \begin{cases} N_1(t) & \text{if } V_1 < c \\
0 & \text{if } V_1 \geq c 
\end{cases}.$$
($V_1$ does not depend on $t$ in your assumptions, there's only one random variable $V_1$.)
Thus there is a positive probability that the trajectory of $N_3$ is identically equal to $0$, whereas this almost never occur for a Poisson process.
A possible rephrasement of your question
Maybe you need to rephrase your problem ? We could imagine the following variant. Two sequences $(X_n)$ and $(Y_n)$ of independent Bernoulli random variables are given. For each $n$, we color the $n$-th event of $N_1$ in blue if $X_n=1$, in red if $X_n=0$. Then $N_3$ is the Poisson process defined by the blue events. We do the similar construction for $N_4$ with $(Y_n)$. Moreover you assume a dependence between $X_n$ and $Y_n$. @NeilG's construction does not match your problem because this dependence appears nowhere in his answer; you clearly start with a construction based on a dependent "coloring".
I have not tried to assess the distribution of $N_3+N_4$ with the construction above. 
Showing the dependency between $N_3$ and $N_4$
Clearly, @NeilG's answer cannot be correct because it does not take into account the dependence assumption (actually his strangely upvoted answer has nothing to do with your question). 
So let's have a first look at my rephrasement of your question and let's show that $N_3$ and $N_4$ are dependent. Denote by $T_1$, $T_2$, $\ldots$ the jumping times of $N_1$ and by $T'_1$, $T'_2$, $\ldots$ the jumping times of $N_2$. 
With my notations above, the first jumping time of $N_3$ is $T_S$ where $S=\min\{n\geq 1 \mid X_n=1\}$ and similarly  the first jumping time of $N_3$ is $T'_{S'}$ where $S'=\min\{n\geq 1 \mid Y_n=1\}$. 
Let's have a look at the joint distribution of $T_S$ and $T'_{S'}$:
$$\Pr(T_S \in \mathrm{d}x, T'_{S'} \in \mathrm{d}x') = \sum_{k,k'} \Pr(T_k \in \mathrm{d}x, T'_{k'} \in \mathrm{d}x') \Pr(S=k,S'=k').$$ 
One has $\Pr(T_k \in \mathrm{d}x, T'_{k'} \in \mathrm{d}x')=\Pr(T_k \in \mathrm{d}x) \Pr(T'_{k'} \in \mathrm{d}x')$ but $\Pr(S=k,S'=k') \neq \Pr(S=k)\Pr(S'=k')$ in general by your dependence assumption. 
I don't continue the calculation, but we see that the equality $\Pr(T_S \in \mathrm{d}x, T'_{S'} \in \mathrm{d}x') = \Pr(T_S \in \mathrm{d}x) \Pr(T'_{S'} \in \mathrm{d}x')$ is not expected, which implies that $N_3$ and $N_4$ are not independent.   
A: The process $N_3(t) + N_4(t)$ cannot be a Poisson process if $N_3(t)$ and $N_4(t)$ have nonzero correlation. The variance of the sum of two random variables is given by $\sigma^2_{X+Y} = \sigma^2_X + \sigma^2_Y + 2\sigma_{XY}.$ But the variance of the number of events occurring in time $t$ in a Poisson process must be equal to the mean since the number of events is a Poisson random variable and the Poisson distribution has $\sigma^2 = \mu.$ 
So if $N_3(t)$ and $N_4(t)$ have positive correlation, then the variance of their sum will be higher than that for a Poisson process with the same mean. Correspondingly, if they have negative correlation, the variance will be smaller than that for a Poisson process. 
Generally the sum does approach a Poisson process as the two component processes approach independence. An interesting case would be if they are uncorrelated but not independent. In that case we do know the first two moments would match. 
The dependence issue was discussed in an earlier question: Dependent thinning Poisson process 
A: I am assuming that we can just replace $v_1, v_2, c$ with a Bernoulli random variable $B$ with probability $p$, which we use to accept events since $v_1$ and $v_2$ are almost surely not sampled at the same time.
The superposition of $N_3$ and $N_4$ are a Poisson process and are equal to $N_5$ by the colouring and superposition theorems.
Imagine a single Poisson process with rate $T = \lambda_1 + \lambda_2$.  Colour each event red with probability $\frac{p\lambda_1}{T}$, green with probability $\frac{p\lambda_2}{T}$, and blue otherwise.  Then, the red events are $N_3$ and the green events are $N_4$.  This construction matches yours, yet in this one it is clear that $N_3$ and $N_4$ are independent (contrary to your statement).
The superposition of independent Poisson processes is a Poisson process.
