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Re: My private key
From: "Thor (Hammer of God)" <Thor () hammerofgod com>
Date: Sat, 12 Jun 2010 17:10:59 +0000

You're killin me over here ;)

And while funny, you actually do raise a good point - I should not use the term "totally secure" like that.  Rather, I 
should say that the encryption mechanisms used are based on industry standards and accepted mechanisms for strong 
encryption:  RSA2048 asymmetric, AES256 symmetric, and SHA256.  

I should have a full work up by the end of the weekend.

t

-----Original Message-----
From: musnt live [mailto:musntlive () gmail com]
Sent: Saturday, June 12, 2010 9:09 AM
To: Thor (Hammer of God)
Cc: Benji; Larry Seltzer; full-disclosure () lists grok org uk
Subject: Re: [Full-disclosure] My private key

On Sat, Jun 12, 2010 at 10:55 AM, Thor (Hammer of God)
<Thor () hammerofgod com> wrote:

It's totally portable, totally secure,

Hello Full Disclosure, I'd like to warn you about "totally secure" and rubber
hose cryptography. While Thor's bold statement of totally secure is so to say
potential and possible the interrogators at Camp X-Ray beg to differ. Yes list
"creative questioning" can yield Thor or anyone else's key and can be
mathematically proving using a patended Craig S. Wright algorithm:

Let P(n) be the statement that says that key+password+...+n = (n/2)(n+1)

Firstly P(n) has to be checked for n=N, which is impossible

It cannot be shown that the truth of P(k-1) implies the truth of P(k).
Because, P(k-1) is the statement key+password+...+(k-1) = ((k-1)/2)k, which is
assumed to be true for k greater than or equal to 2 however N cannot be
calculated.

Next add k to both sides of statement P(k-1) to get
key+password+...+(k-1)+k = ((k-1)/2)k+k. Taking out a factor of k on
the right hand side of the equation leaves key+password+...+k = (((k-
1)/2)+1)k = k((k/2) + (1/2)) =(k/2)(k+1), which implies that P(k) is true.
Condition 2 has been satisfied.

Both conditions of the statement for the principle of mathematical induction
have been satisfied but N is never established and the proof is inconclusive, in
other words P(n) is true for all positive integers n and nothing more given
that: B(eer)||T(orture)||M(oney) trump all
so:

B+M=P(*) || T=P(*)

Please contact Mr. Wright LLC, PhD, DDS, CISSP, GSE, GSE, GSE for future risk
metrics. Did forget I mention GSE?

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