if a hash is difference-unpredictable, will it keep unpredictable given an offset? Let $H(k,m)$ an universal hash function that is difference-unpredictable.

Let $\mathcal{A}$ be an attacker who sends tuples $(m_0, m_1, \delta)$, and $\mathcal{A}$ wins the game if one of the tuples satisfy $H(k,m_1) -H(k,m_0) = \delta$.

$H$ will be difference-unpredictable when the chance of $\mathcal{A}$ win is negligible.

Suppose $H$ is difference-unpredictable, how can I show that the function $H'((k,x),m) = H(k,m) + x$ will also be?

Both functions' image is all naturals smaller than a given prime $p$, one can think of the functions like $H(k,m) = (...)\bmod p$. 



I think the proof remains on this argument:


the attacker chooses $(m_0, m_1, \delta)$
let $H(k, m_0) = h_0$ and $H(k, m_1) = h_1$
let $d$ be the "wrapping" distance from $h_0$ to $h_1$, that is:


$d = h_1 - h_0\;$ if $\;h_0 \leq h_1$
$d = p - h_0 + h_1\;$ if $\;h_0 &gt; h_1$



If $H'$ only offsets $h_0$ and $h_1$ by $x$, the "wrapping" distance will be the same and it will give no better chance for the attacker to guess a difference change in the results and his chance will remain negligible.
What's $H'((k, x), m_0) - H'((k, x), m_1)$?