FFXIV riddle

There are 9 different letters in THAUMATURG and METEOR

THAUMRGEO and we have the numbers from 1 to 9

The equation is THAUM + ATURG = METEOR

I.) Even if we chose the highest possible 5-digit number that uses each digit only once for both words (98765)
the result would still be 197530 so we can safely assume that the word METEOR starts with a 1.

==> M = 1

II.) The next part I want to point out is more apparent if I write it like this:

     |
     V
    THAUM
   +ATURG
  =METEOR

In this row you see that H + T = T. This can only occur under two conditions, H has to be 0 or 10. Neither 0 nor 10
are in our list but we have access to a 9 and we can add a 1 if the previous addition "A+U" was bigger than 10.
We end up with "H + T = 10+T". "10+T" means the digit we want for T and a carry.

So we now learned that:

 M = 1, H = 9 and A+U>10

III.) Actually, A+U has to be even bigger than 11 because each digit may only occur once and 1 is already used (by M),
so A+U has to be bigger than 11

M = 1, H = 9 and A+U>11

IV.) Alright, we learned above (II.) that H + T is bigger than 10 so we have to add 1 to the addition on the far left.
We now have:

 T + A + 1 = 10+E

Let's look for another E

A + U = E (and we know from above, A+U>11)

This time there may NOT be a carry in this addition. If we were to add the extra 1 we would end up with

A + U + 1 = 10+E

considering that  T + A + 1 = 10+E
this would mean that T = U and this can not be the case. Each digit may only be used once. So it HAS TO BE

A + U = 10+E

And it has to be 10+E on the right side because we need that carryover 1 for H+T.

V.) Let's continue on the far right. We have M + G = R. We know that M = 1 so the equation is

1 + G = R. There is not much we can do here other than note that since we have no 0 and we are adding G to 1, R needs to be
at least 3. 1 is already used so the next smaller digit for G is 2, so R has to be 3 or bigger.

VI.) Next part, U + R = O. This result has to be smaller than 10 since we may not have a carryover (this would contradict IV.).
We also already used 9 (for H) so to stay smaller than 10 we need U + R < 9 and we know (from V.) that R >= 3 so U has to be
smaller or equal to 5. If U were bigger than 5 we would end up with U+R > 8 and this is not possible.

VII.) Ok let's get to the final parts. We now have basically everything we need:

M = 1, H = 9, T+A+1=10+E, A+U=10+E, O = U+R < 9, R >= 3, U <= 5, G >= 2

The most important parts are these two equations:
  T + A + 1 = 10+E
  A + U     = 10+E

Solving them for E yields:
 A + T - 9 = E
 A + U - 10 = E

We can instead write:
 A + T - 9 = A + U - 10

 And simplify it:
 T + 1 = U

So U is always 1 bigger than T, but how can we use this? We know that U <= 5 so we now also know that T <= 4.

------------------------------------------------------
Testing all possible numbers from here on:

  THAUM
 +ATURG
=METEOR

M = 1, H = 9. Let's try G = 2  => R = 3

  T9AU1
 +ATU32
=1ETEO3

Ok, how about U = 5 => O = 8

  T9A51
 +AT532
=1ETE83

At this point we run into a problem. We need "A+5 = E" to be bigger than 10. We only have 6 and 7 left to use for A but
if we use 6 or 7 we get 6 + 5  = 10 + 1 (or 7 + 5 = 10 + 2), so we would end up with E=1 or E=2. Both are not possible since
we already used these numbers.

Let's try a different number for "G":

M = 1, H = 9. Let's try G = 3  => R = 4

  T9AU1
 +ATU43
=1ETEO4

Ok, we know that U+R=O has to be smaller than 9 so we can only chose U=2 but this already causes a problem since U = T+1.
U=2 would mean that T=1 but 1 is already used. So this does not work.

Again, let's try a different number for "G":

M = 1, H = 9. Let's try G = 4  => R = 5

  T9AU1
 +ATU54
=1ETEO5

Right, U+5<9 so U<4. Since U=T+1 and 1 already being used U can't be 2 so U needs to be 3.

  T9A31
 +AT354
=1ETE85

Once again "A+3=E" needs to be bigger than 10. So A has to be 8 or 9 but both are already used ... tough luck.

Next number for G:

M = 1, H = 9. Let's try G = 5  => R = 6

  T9AU1
 +ATU65
=1ETEO6

Welp, now U has to be 2 because U+6<9, but it can't be because T would be 1 which is already used.

The next one G = 6 ends even quicker:

M = 1, H = 9. Let's try G = 6  => R = 7

  T9AU1
 +ATU76
=1ETEO7

U+7 < 9 but 1 is already used. Done, no solution for this problem.