Let:
- $T_{p,r}(n,x)$ be the count of integer $i$ such that:
- $p$ is an odd prime
- $n,r<p,x$ are integers
- $n - x \le i < n$
- $i \equiv r \pmod {p}$
- for all prime $q > 2$ where $i \equiv r \pmod {q}$, it follows that $p \le q$
Does it now follow that for any odd prime $p$, any integers $r, n, x$:
$$T_{p,1}(0,x) \ge T_{p,r}(n,x) \ge T_{p,0}(0,x)$$
I am finding it difficult to prove that this is the case in light of the 5th bullet point which is my attempt to generalize the concept of least prime factor to modular arithmetic.
Am I wrong in my assumption?
Edit: Attempting an argument for $T_{p,1}(0,x) \ge T_{p,r}(n,x)$ where $r=1$:
It occurs to me that this special case of my assumption is straight forward to show:
(1) Assume it is not true so that there exists $p,n,x$ such that:
$$T_{p,1}(0,x) < T_{p,1}(n,x)$$
(2) Let:
$S_{0,x,q}$ be the set of integers $i$ where $-x \le i < 0$ and there exists a minimal prime $q$ such that $q < p$ and $i \equiv 1 \pmod q$ and $i \equiv 1 \pmod p$
$S_{n,x,q}$ be the set of integers $j$ where $n-x \le j < n$ and there exists a prime $q$ such that $q < p$ and $j \equiv 1 \pmod q$ and $j \equiv 1 \pmod p$
$|S_{0,x,q}|$ be the count of elements in $S_{0,x,q}$ and $|S_{n,x,q}|$ be the count of elements in $S_{n,x,q}$
(3) There must exist at least one $q$ where $m=|S_{0,x,q}| > |S_{n,x,q}|$
(4) But then, $x \ge mpq+1$ Otherwise, it would be impossible for $m$ such elements to exist.
(5) It follows $n-x, n-x+1, n-x+2, \dots, n-x+mpq$ forms a complete residue system modulo $n-x+mpq+1$
(6) But then $|S_{n,x,q}| \ge m$ which contradicts step(3).
(7) So, we can reject our assumption at step(1).
Edit 2: Attempting an argument for $T_{p,1}(0,x) \ge T_{p,r}(n,x)$ where $r>1$:
(1) Assume it is not true so that there exists $p,n,x$ such that:
$$T_{p,1}(0,x) < T_{p,1}(n,x)$$
(2) From the argument above, we can assume that $x \ge pqm+1$ and $x < pqm+r$. Otherwise, the same argument applies.
(3) We can assume that:
$$\sum_{\text{all }q}|S_{0,x,q}| = \sum_{\text{all }q}|S_{n,x,q}|+1$$
Assume that $\sum\limits_{\text{all }q}|S_{0,x,q}| > \sum\limits_{\text{all }q}|S_{n,x,q}|+1$
Then, there exists primes $q_1, q_2$ such that $q_1p +1 < q_1p + p +1 \le x < q_1p + r$
But this is impossible since by assumption $r < p$
(4) Let $U_{p,r}(n,x)$ be the count of integers $i$ where $n-x \le i < n$ and $i \equiv r \pmod p$ so that:
$$T_{p,r}(n,x) = U_{p,r}(n,x) - \sum_{\text{all }q}|S_{n,x,q}|$$
(5) It follows from step(1) that $U_{p,1}(0,x) = mpq$ while $U_{p,r}(n,x) = mpq-1$
(6) Combining step(5) and step(3):
$$T_{p,1}(0,x) = U_{p,r}(n,x)+1 - \sum\limits_{\text{all }q}|S_{n,x,q}|-1 = T_{p,r}(n,x)$$
(7) This contradicts step (1) so we can reject our assumption in step(1).
Note with your definition of $T_{p,r}(n,x)$, your fourth & fifth bullet points combined basically state you're looking for where the least odd prime factor (I'm calling call it $\operatorname{lopf}$ for simplicity) of $r + (-i)$ is $p$.
Your conjecture is equivalent to all of the following $3$ inequalities being true:
$$T_{p,1}(0,x) \ge T_{p,r}(n,x) \tag{1}\label{eq1A}$$
$$T_{p,r}(n,x) \ge T_{p,0}(0,x) \tag{2}\label{eq2A}$$
$$T_{p,1}(0,x) \ge T_{p,0}(0,x) \tag{3}\label{eq3A}$$
Of course, if the first $2$ are true, then so is the third one. However, a counter-example to \eqref{eq1A} is $T_{3,1}(0,1) = 0$, but $T_{3,1}(5,1) = 1$. Also, a counter-example to \eqref{eq2A} is $T_{5,0}(16,5) = 0$ (since with $q = 3$, then $q \mid 15$ and $q \lt p$, so $15$ is excluded), but $T_{5,0}(0,5) = 1$. Nonetheless, \eqref{eq3A} is always true, as I show later.
Regarding your Edit attempt to prove $T_{p,1}(0,x) \ge T_{p,r}(n,x)$ where $r = 1$, this is a special case of \eqref{eq1A}, and I already gave a counter-example with $r = 1$, i.e., with $p = 3$, $x = 1$ and $n = 5$. In your attempted proof by contradiction, with your $(3)$, you seem to be assuming that if the number of integers where the $\operatorname{lopf}$ is $p$ in one range is greater than in a second range, the second range must have at least one prime $q \lt p$ (you don't state your $q$ must be odd, but it only makes sense since your $T_{p,r}(n,x)$ definition only worries about odd prime factors) where the number of values with $\operatorname{lopf}$ being $q$ is greater than that in the first range. However, as my example shows, this is not always the case. For one thing, if $p = 3$, there is not even any such $q \lt p$. However, even if $p \gt 3$, then it's still not necessarily true, e.g., with $T_{5,1}(0,1) = 0$ and $T_{5,1}(7,1) = 1$.
Your $(4)$ states $x \ge mpq+1$. However, with some $q \lt p$, have $i = -x = -pq + 1$. Then $i \equiv 1 \pmod{p}$ and $i \equiv 1 \pmod{q}$, so $m = 1$. However, $x = pq - 1$. The correct statement to make is actually $x \ge mpq - 1$.
This, of course, affects your $(5)$. Also, you state it forms a complete residue system modulo $n - x + mpq + 1$. However, for example if $n = 0$ and $x = mpq + 1$, this modulo expression is $0$. I believe you meant to state it is for modulo $mpq + 1$ instead. Nonetheless, a minor point is you actually want to instead state the range of integers contains a complete residue system modulo $mpq$, to then try to conclude your $(6)$. However, even then, there's a potential complication if any of those values which have a factor of $pq$ also have an odd prime factor smaller than $q$.
With your Edit 2 attempt to prove $T_{p,1}(0,x) \ge T_{p,r}(n,x)$ where $r \gt 1$, a specific counter-example is $T_{3,1}(0,1) = 0$ and $T_{3,2}(6,1) = 1$. Since the rest of your argument is based on the faulty logic and results of the first edit proof, this proof likewise is not reliable.
To prove \eqref{eq3A}, first here is a definition of the $\operatorname{lopf}$ function, where $m \in \mathbb{Z}$ and $\mathbb{P}$ is the set of all primes:
$$\operatorname{lopf}(m) = \begin{cases} 1 & m = \pm 2^j, j \in \mathbb{Z}, j \ge 0 \\ \min\{q : q \mid m, \, q \in \mathbb{P}, \, q \gt 2 \} & \text{otherwise} \end{cases} \tag{4}\label{eq4A}$$
Thus, your $T_{p,r}(n,x)$ can be defined as
$$T_{p,r}(n,x) = \left|\{i : \operatorname{lopf}(r - i) = p, \, n - x \le i \lt n \}\right| \tag{5}\label{eq5A}$$
As $x$ increases starting from $1$, both sides of \eqref{eq3A} stay at $0$ until $x$ gets to the first value where $\operatorname{lopf}(x + 1) = \operatorname{lopf}(-x - 1) = p$. In that case, since $-x \equiv 1 \pmod{p}$, then $T_{p,1}(0,x) = 1$, but $T_{p,0}(0,x) = 0$ still. However, when $x$ increases by $1$ more, then $\operatorname{lopf}(x) = p$, so $T_{p,0}(0,x) = 1$, which means both sides are equal. This continues until the next larger $x$ where $\operatorname{lopf}(x + 1) = p$, with the pattern repeating, i.e., the left side becomes $1$ greater then, but becomes equal with the next larger $x$. In general, whenever $\operatorname{lopf}(x + 1) = p$, then $T_{p,1}(0,x) = T_{p,0}(0,x) + 1$ and, otherwise, $T_{p,1}(0,x) = T_{p,0}(0,x)$.