I do not know at what stage you are in dealing with linear recurrences with constant coefficients, and in particular whether you already have general tools.
If you do not yet have general tools for recurrences of this kind, and even if you do, it may be useful to compute a few more terms, in order to get additional insight about the sequence $(t_n)$.
We have $t_4=t_3+(t_1-t_0)=3+(1-0)=4$.
Also, $t_5=t_4+(t_2-t_1)=5$.
Also, $t_6=t_5+(t_3-t_2)=6$.
Also, $t_7=t_6 +(t_4-t_3)=7.$
The pattern is hard to miss! Now let us prove that the pattern always holds. An easy way to do this is to define the sequence $(s_n)$ by $s_n=n$. We want to show that $t_n=s_n$ for all $n$.
It is easy to verify that $s_n=s_{n-1}+s_{n-3}-s_{n-4}$ for all $n$, since $n-1+n-3-(n-4)=n$. Also, $s_n=n$ for $n=0, 1, 2, 3$. Of course.
So the sequence $(s_n)$ obeys the same recurrence as $(t_n)$, and the same initial conditions. The two sequences are therefore identical.
Or else we can do a formal proof by induction that $t_n=n$ for all $n$.
The induction step is as follows. Suppose that $t_k=k$ for all $k. We want to show that $t_n=n$. We are given that $t_n=t_{n-1}+t_{n-3}-t_{n-4}.$ By induction hypothesis, $t_{n-1}=n-1$, $t_{n-3}=n-3$, and $t_{n-4}=n-4$. Calculate. We get that $t_n=n-1+ n-3 -(n-4)=n$. Note that this is fundamentally the same argument as the first one.