Update: Replaced the letter $E$ with $F$ since wolframalpha gets confused with $E$.
Now you can ask wolfram to solve this, so that the theory can be applied to a larger solution space, supplying the general solution; but then it has to be used to zone in on the specific 'boundary conditions' of interest.
Consider
SPECIFICATION:
$\tag 1 8000 F + 990 A + 100 D + 99791 = 10000 H + 10100 M + 2 S + Y + 1000 X$
where
$\tag 2 F,A,D,H,M,S,Y,X \in \{0,2,3,4,5,6,7,8,9\}$ and $\tag 3 \text{The set } \{F,A,D,H,M,S,Y,X\} \text{ contains } 8 \text{ elements}$
This came about after trying to solve a problem in recreational math; see
Christmas Cryptarithm: “HERES+MERRY+XMAS=READER”
I am hoping that by presenting the problem in this way it is more amenable to the application of mathematical theories, such as number theory.
Working with wolframalpha number theory engine over all the integers, on the 'high level logic breakdown' I focused on two cases
MAIN CASE 1:
$\quad 2 S + Y = 11$
$\quad A = 8$
$\quad \text{Solve } 8000 F + 990 * 8 + 100 D + 99791 = 10000 H + 10100 M + 11 + 1000 X$
$\text{XOR}$
MAIN CASE 2:
$\quad 2 S + Y = 21$
$\quad A = 7$
$\quad \text{Solve } 8000 F + 990 * 7 + 100 D + 99791 = 10000 H + 10100 M + 21 + 1000 X$
Main Case 1 Work
I know that MAIN CASE 1 has more than one solution but at least one of them can be retrofitted for a $S$/$Y$ solution giving an answer to the SPECIFICATION.
Here is the solution that can't 'accommodate' $S$ and $Y$:
$\quad F = 2$
$\quad A = 8$ (already given)
$\quad D = 0$
$\quad H = 5$
$\quad M = 7$
$\quad X = 3$
Here is the Christmas Cryptarithm solution for Main Case 1 that solves the SPECIFICATION:
$\quad F = 4$
$\quad A = 8$
$\quad D = 0$
$\quad H = 6$
$\quad M = 7$
$\quad S = 3$
$\quad Y = 5$
$\quad X = 9$
Main Case 2 Work
I suspect that MAIN CASE 2 has more than one solution but at least one of them can be retrofitted for a $S$/$Y$ solution giving an answer to the SPECIFICATION.
Here is the solution for Main Case 2 that solves the SPECIFICATION:
$\quad F = 3$
$\quad A = 7$
$\quad D = 2$
$\quad H = 4$
$\quad M = 9$
$\quad S = 8$
$\quad Y = 5$
$\quad X = 0$
Problem: Show that there is exactly one solution to the SPECIFICATION for Main Case 1 and exactly one solution to the SPECIFICATION for Main Case 2.
Note that if we allow $X$ to be $0$ in the Christmas Cryptarithm we get a second solution:
$\quad 43138 + 93115 + 0978 = 137231$
