Abstract Rewriting Challenge (Robbers) This is a somewhat proof-golf-like cops-and-robbers challenge. This is the robbers' thread; the cops' thread is here.

Robbers

The cops will post abstract rewriting systems. Your task is to crack their submissions by proving that the target string either can or can not be reached from the source string by applying their rewrite rules. (You can do this either by posting a sequence of rewrite rules that starts with the source string and ends with the target, or by proving mathematically that this does, or does not, exist.)

See the cops' thread for details and definitions.
jimmy23013

Let's work backwards for this one. We first turn the digits into their binary representations. We go from VW626206555675126212043640270477001760465526277571600601 to VW++__+_++__+____++_+_++_++_+++_++++_+__+_+_++__+___+_+____+___++++_+______+_+++___+__++++++________++++++____+__++_+_++_+_+_++__+_+++++++_++++__+++_______++______+. Next, we keep applying the inverse of DCW:W+ and DW:W_ until we clear all symbols. Our result is now VDCDCDDDCDDCDCDDDCDDDDDCDCDDCDDCDCDDCDCDDCDCDCDDCDCDCDCDDCDDDCDDCDDCDCDDDCDDDDCDDCDDDDDCDDDDCDCDCDCDDCDDDDDDDCDDCDCDCDDDDCDDDCDCDCDCDCDCDDDDDDDDDCDCDCDCDCDCDDDDDCDDDCDCDDCDDCDCDDCDDCDDCDCDDDCDDCDCDCDCDCDCDCDDCDCDCDCDDDCDCDCDDDDDDDDCDCDDDDDDDCW. We now want to make this string match VD+C+W; that is, we want to move all of the Ds to the left of all of the Cs. This can be done by reversing DCC:CD. We do this by repeating the following algorithm:


Find the first D that is to the right of a block of Cs.
Move the D to the left of that block.
Double the number of Cs.


Through some math, we can determine that we will end up with 123 Ds and 4638704741628490670592103344196019722536654143873 Cs (you were right about this not fitting in an SE answer... I doubt this would fit if stored as states of all of the atoms in the Earth combined :P).

If we keep applying the reverse of V:VD, we can get rid of all of the Ds now, so we get VCCC.......CCCW. We convert the V back into YZ. Now we have YZCCC.......CCCW.

We want to be able to get rid of all of the Cs and have it in the form YAAA...AAABBB...BBBZW. Fortunately, this can be done by the following method. Firstly, we inverse-apply YB:Y 587912508217580921743211 times to get YBBB.......BBBZCCC.......CCCW. Then, we repeat the following sequence of steps (where [?*] means any number of ?, not necessarily greater than zero):


Inverse-apply CZ:ZC 587912508217580921743211 times to get Y[A*]BBB.......BBBCCC.......CCCZCCC.......CCCW
Inverse-apply CB:BC many times to get Y[A*]BCBCBC.......BCBCBCZCCC.......CCCW
Inverse-apply AZ:Z and AB:BCA many times to get Y[A*]ABBB.......BBBZCCC.......CCCW


Through induction, we see that we can move the BZ combination all the way to the end (except before the W) and then the number of As is 1/587912508217580921743211 of the number of Cs, leaving us with 7890127658096618386747843 As. We now have YAAA.......AAABBB.......BBBZW. Convert the ZW back to a U, then inverse-apply U:BU many times to keep only 2 Bs and then convert the BBU to a T, and you now have YAAA.......AAAT. Then, you can inverse-apply T:AAAAAT many times to get YAAAT because the number of As was 3 larger than a multiple of 5.

Thanks for the challenge!