Once more, sorry for my poor English.

Your English is much better than my Russian. :-) I've cleaned up spellings and punctuation in the quote below, to reflect my understanding. Was I correct?

I want to say that DES isn`t group. We can`t say that: A k1, k2; E k0; ST des(k2, des(k1, text))=des(k0, text) (A means "for all", E means "exist", ST means "such that");

or: E k1, k2; A k0; ST des(k2, des(k1, text))!=des(k0, text)

I say that: E f(k, text); A k1, k2; E k0 ST len(k0)=len(k1)=len(k2) and des(k2, des(k1, text))=f(k0, text)

f() is like des() but f() is not des()!!! My friend says that the intruder must try only 2^56 keys for des(des()).

What's missing in the last statement is the qualification of <text>. Is it A text or E text ? Meanwhile, some method of learning f would be nice :-). Is it also claimed that: E g(k,text); A k1,k2,text; E k0 ST len(k0)=len(k1)=len(k2) and des(k2, f(k1, text)) = g(k0,text) for f(,) as defined above? (That is, does this method continue to apply for an indefinite number of DES copies?) - Carl