The 40th document begins with "Till whatsoever star that guides my moving". What is the first word in Document 39? It must be in the table of letterword relationships. So, it must be derived from the first letter of Doc. 38. Let us call this letter Y. So, the first word of Doc. 38, let's call this "X", must contain Y. Now, where does it contain Y? If Y is the first letter of "X", then "Till" comes from Y. But "X" comes from Y too, so we have "X"="Till". But this is impossible, as "whatsoever" cannot be derived from i, because it doesn't contain i. By similar reasoning, we find that Y cannot be any of the first 7 letters of "X". So, we are forced to conclude that "X" is 8 or more letters long and that it has Y in the 8th place or after that. We now know that "X" is the first word of Document 39. So, Y is the first letter of Document 38. In other words, Doc. 38 begins with a word whose first letter is Y. This word is not "X" because we have proved that "X" has Y in the 8th place or after that. Thus, Document 37 does not begin with Y.
We now need to stop and prove something else. We need to prove that if any document contains a letter (say Z), then Document 26 contains that letter, and so do all documents after it. Let S(n) be the set of letters that compose Document n. For example, if Document 1 was "apple", then S(n) = {a,p,l,e}. We define a "new letter" to be a letter that has appeared in Document X but not in Document X1. (Any letter in Doc. X1 will appear in Doc. X because the word for letter Z contains the letter Z, remember.) Also note that S(n+1) can be found if S(n) is given (and the relationships have been determined) because every letter in Doc. n will be changed to its word in Doc. n+1, and repeating of the word will not affect the set. Then, we can find that is S(n+1) = S(n), then all future sets will be identical too. Therefore, if Document n has any new letters, then all documents before it must have at least one new letter too. So, assume, contrary to what is to be proved, that Doc. 27 has a new letter. Then all previous ones obviously have one too. Then S(27) has less than or equal to 26 elements, S(26) has less than or equal to 25, and so on, to S(1) has less than or equal to 0 elements, which is impossible, so Document 27 cannot have any new letters.
Back to the problem: In Document 39, we know we have the letter Y. Thus, in Document 26 and all later ones, we also have at least one occurence of letter Y, including Document 37. Since Document 38 begins with Y, and this Y wasn't caused by a Y in Doc. 37, Document 37's Y will cause another Y elsewhere in Document 38. So, "X" will occur somewhere else in Document 39, for 2 "X"'s in Document 39. At the beginning, we have shown that an "X" in Document 39 yields a sentence beginning with "Till whatsoever star that guides my moving". So, two "X"'s mean two "Till whatsoever star that guides my moving"'s.
Q.E.D. _________________ This statement is false. Conclusion: "This statement is false" is not a statement.
