In germany we have very hard laws for CP. The form of CP is free even writing a CP Text is forbidden or having a encrypted code. The point is not the technic, olny the sexual aspect of minors.
So yes , a windows that would need that CP dll file would be illegal to store.
Thats my point. I think goverment knows about that, maybe some goverment even placed it there.
And when they need to forbid Bitcoin, they do with this argument
Well, the good news is that looking at the paper itself it seems like the researchers didn't even find CP, there was only a claim they found on an online forum that they didn't actually verify for ethical (and possibly legal) reasons:
Bitcoin’s blockchain contains at least eight files with sexual content. While five files only show, describe, or link to mildly pornographic content, we consider the remaining three instances objectionable for almost all jurisdictions: Two of them are backups of link lists to child pornography, containing 274 links to websites, 142 of which refer to Tor hidden services. The remaining instance is an image depicting mild nudity of a young woman. In an online forum this image is claimed to show child pornography, albeit this claim cannot be verified (due to ethical concerns we refrain from providing a citation).
The bad news is that of course that's not stopping anyone from adding CP in the future.
The good news is that it's still unlikely that German courts (or any court, really) would put anyone behind bars for a binary blob of data on the Bitcoin blockchain. Law is based on the legislative intent rather than technicalities and in this case it's about content rather than binary data.
This Heise comment said it best:
pi = 3,141 592 653 589 ...
In den Nachkommastellen soll jede Zahlenfolge die es gibt abgebildet sein.
Damit auch die Zahlenfolge von jedem in Zahlen kodiertes KiPo Bild.
Nur an welcher Stelle sich diese Zahlenfolge befindet muss man selber suchen.
-> also Pi verbieten!
Let's ban the number Pi because it contains every kind of CP imaginable! All you need to do is find the correct starting point in its digits.