TLPTPW - The I just won edition

 Lie-ins are most definitely winning :) but only up to the point they stop you having enough time to do everything you have to once vaguely perpendicular. Which I think *scowl* is how you know you're a grown up. 

CODE SOLUTION: 

ZmNt : the last letter of the alphabet followed by m N t (make NOT true) means you're right Mike. We can ignore Z
ZmRt :  whilst remembering that when then followed by m R t (make REVERSE true)  transfers attention to the first letter of the alphabet - which should be easy enough to find out from a dictionary I reckon. 
ZmNt    )
ZmRt    )
ZmNt    )   and because we now have a letter determined for this segment, these terms cancel one another out
ZmRt    )
ZmNt    )
ZmRt    )
ZwNt        until  w : meaning: whoa => (ie - end cancelling; end first letter, thereby making => w N t (whoa NOT true) which means that the first letter is not a standalone one.
     

THEREFORE - via standard hiatical reasoning
ZmNt   )
ZmRt   )   we must cancel initial redundancies prior to a code line duplicated sequentially
ZmNt   )
ZmRt   )
ZmNt        as this one is here
ZmNt        with this one
ZmRt   )
ZmNt   )  then cancel a reversed redundancy (as here) to preserve the doubled uppercase metric in its lowercase form, giving us two n's
ZwNt ............... (see up there somewhere - but change first to second)
     

THEN - obviously, as we've just cancelled a reversed redundancy, we have to follow the laws governing cognitive semantic resonance patterning as prescribed by *stare straight ahead* a percussive triangle in its struck state. So!Looking at the next segment of Jonathan's code: ZmNt : ZmNt : ZmRt : ZmNt : ZmNt : ZmNt  : ZmNt  : ZmNt after stripping out some tactical buffering gives us the word llilllllly .... which might look problematic until we remember that it's only its vowels that will have any significance with regard the code... thereby  giving us an... i.

The nebulousness of y in its role at the end of a word is generally discounted as a y at this stage and replaced with its phonetic equivalent, ie: e to give us the second letter in this segment - but you could leave doing that until the end unless you're likely to forget to
ZwNt ............... (like I said before up there plussing one)
     

This next segment I expect most everyone already knows the answer to, so I won't go into the proof in too much depth, other than to reiterate that in a 4-column code structure, like here, the last cypher in the fourth segment is the last letter in the first word.
ZmNt
ZmRt
ZmNt
ZmNt  which is t
ZmRt
ZmRt
ZmNt
ZmNt <--- see?
ZwNt .................. again another whoa clause... but do feel free to murmur an impressed wow instead if you wish *wait expectantly*

ZmNt 
ZmNt 
ZmRt 
ZmRt  
ZmNt
ZmNt 
ZmRt
ZmRt  ... nn rr nn rr is the sequencing here and also in the following segment. 
ZwNt ..................

ZmNt  It really is the most fascinating part of the puzzle because of its reiteration of the reversed n's to become u's 
ZmNt
ZmRt 
ZmRt
ZmNt
ZmNt  
ZmRt
ZmRt
ZwNt......................  thereby reinforcing the electrostigmatic field defining double u as uu and then as w which is the first letter of the second word! :)

ZmNt
ZmRt
ZmRt
ZmRt
ZmNt  
ZmRt
ZmNt
ZmNt
ZwNt........................ Then (as amply demonstrated here) o can be the only solution to this segment. It's so round! And being mostly made of an empty space - all forms of reversal have the same end result.
 
ZmNt
ZmNt
ZmRt
ZmNt
ZmNt
ZmNt
ZmNt  and with the surfeit of n's here ... n is the last letter.
ZmR 

And there you have it! :) Solved!!

 I have chronic constant megalomania, thnak you very much! It's far superior to the other types. The megalomaniac's megalomania as it were.

*lower light flaps... then release* Well thnak you for confirming that for us, Mike. I hope the above didn't overly disturb your equilibrium?

*Beam happily at Einsteinians everywhere* Hope you're all having a lovely weekend :)

Well obviously, stands out a mile doesn't it??

Almost to the point it went without saying, yes.

:)

I see you had a look in the dictionary, Chris...

edit: ... and then branched out adventurously! :)

ZmNtZmRtZmNtZmRtZmNtZmRtZmNtZmRtZwNt
ZmNtZmRtZmNtZmRtZmNtZmNtZmRtZmNtZwNt
ZmNtZmNtZmRtZmNtZmNtZmNtZmNtZmNtZwNt
ZmNtZmRtZmNtZmNtZmRtZmRtZmNtZmNtZwNt
ZmNtZmNtZmRtZmRtZmNtZmNtZmRtZmRtZwNt
ZmNtZmNtZmRtZmRtZmNtZmNtZmRtZmRtZwNt
ZmNtZmRtZmRtZmRtZmNtZmRtZmNtZmNtZwNt
ZmNtZmNtZmRtZmNtZmNtZmNtZmNtZmR

This is the problem one has with messages, one has to figure out the matrix before you can begin to decode.

As you see above the "w"'s align.

Going again, one notices the only other change is either an "R" or an "N"

Assigning 0 to R and 1 to N, or course that might be reverse

10101010        01010101  U
10101101        01010010  R
11011111        00100000  (space)
10110011        01001100  L
11001100        00110011  3
11001100        00110011  3
10001011        01110100  t
11011110        00100001  !

 decide if ASCII or EBCDIC and read message

Me thinks I ought to clean my glasses.  It might make Annie's solution make more sense.  :0)

Oh. Erm... I wouldn't actually count on that, Kathryn. No.

;)

I did have something resembling Gary's method at one point *methodically scan floor for piece of paper* although I can't remember whether R was 1. Whichever way round it was. these: â £  ã ¢ £ @ @ ñ (jumbled so as not to set off any world wide panic) were not quite what I was expecting after the earlier string of abuse that came out of the converter table I was using.  

 No. So I filed it. The piece of paper I mean. It's just as well I did or I might never have broken the code! :)

I don't know about anyone else, but I'm getting the feeling I haven't won yet...

unless I've tripped up, simply by not putting it into quotes...

Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh! Oh that's very good! I'll be back. 

:)