Given an arbitrary block C$1,$ choose C$2$ so that RSAH$($C$1,$ C$2)=$ RSAH$($B$1,$ B$2).$ Thus, the hash function does not satisfy weak collision resistance. $11\mathrm{.}6$ Suppose H$($m$)$ is a collision$-$resistant hash function that maps a message of arbitrary bit length into an n$-$bit hash value. Is it true that, for all messages x$,$ x$$ with x $!=$ x$,$ we have H$($x$)!=$ H$($x$)$ Explain your answer.