Pratyya and Payel have a number each, $n$ and $m$ respectively, where $n>m.$ Everyday, Pratyya multiplies his number by $2$ and then subtracts $2$ from it, and Payel multiplies his number by $2$ and then add $2$ to it. In other words, on the first day their numbers will be $(2n-2)$ and $(2m+2)$ respectively. Find minimum integer $x$ with proof such that if $n-m\geq x,$ then Pratyya's number will be larger than Payel's number everyday.
1. Define the initial numbers as \( n \) and \( m \) with \( n > m \). Let \( d_0 = n - m \) be the initial difference between Pratyya's and Payel's numbers.

2. On the first day, Pratyya's number becomes \( 2n - 2 \) and Payel's number becomes \( 2m + 2 \). The new difference \( d_1 \) between their numbers is:
   \[
   d_1 = (2n - 2) - (2m + 2) = 2n - 2m - 4 = 2(n - m) - 4 = 2d_0 - 4
   \]

3. Generalize this for any day \( t \). Let \( d_t \) be the difference on day \( t \). Then:
   \[
   d_{t+1} = 2d_t - 4
   \]

4. To ensure that Pratyya's number is always larger than Payel's number, we need \( d_t \geq 0 \) for all \( t \). We need to find the minimum initial difference \( d_0 \) such that \( d_t \geq 0 \) for all \( t \).

5. Consider the recurrence relation \( d_{t+1} = 2d_t - 4 \). We solve this recurrence relation by finding its fixed point. Set \( d_{t+1} = d_t = d \):
   \[
   d = 2d - 4 \implies d - 2d = -4 \implies -d = -4 \implies d = 4
   \]

6. The fixed point \( d = 4 \) suggests that if \( d_0 = 4 \), then \( d_t \) will remain 4 for all \( t \). However, we need to ensure that \( d_t \geq 0 \) for all \( t \) starting from \( d_0 \).

7. To find the minimum \( d_0 \), we consider the series:
   \[
   d_t = 2^t d_0 - 4(2^t - 1)
   \]
   For \( d_t \geq 0 \):
   \[
   2^t d_0 - 4(2^t - 1) \geq 0 \implies 2^t d_0 \geq 4(2^t - 1) \implies d_0 \geq 4 \left(1 - \frac{1}{2^t}\right)
   \]

8. As \( t \to \infty \), \( \frac{1}{2^t} \to 0 \), so:
   \[
   d_0 \geq 4
   \]

9. Therefore, the minimum integer \( x \) such that \( n - m \geq x \) ensures Pratyya's number is always larger than Payel's number is \( x = 4 \).

The final answer is \( \boxed{4} \)