What is a Ham Word™?

Question

_{This is in the spirit of the What is a Word™/Phrase™ series started by JLee with a special brand of Phrase™ and Word™ puzzles.}

If a word conforms to a special rule, I call it a Ham Word™.
Use the examples below to find the rule.

Ham Words	Not Ham Words
Bert	Ernie
DAMP	RAMP
towel	towels
GUNS	SWORDS
toxic	healthy
Porto	Lissabon
HAM	BACON
fillet	fish
SPAM	MAIL
roll	wrap
PUNS	PUNCH
troll	goblin
Kiel	Hamburg
MAP	FLATTEN
flow	stream
SPAN	SPLIT

In case you want it in CSV:

Ham Words,Not Ham Words
Bert,Ernie
DAMP,RAMP
towel,towels
GUNS,SWORDS
toxic,healthy
Porto,Lissabon
HAM,BACON
fillet,fish
SPAM,MAIL
roll,wrap
PUNS,PUNCH
troll,goblin
Kiel,Hamburg
MAP,FLATTEN
flow,stream
SPAN,SPLIT

The puzzle relies on the series' inbuilt assumption, that each word can be tested for whether it is a Ham Word™ without relying on the other words.

These are not the only examples of Ham Words™, many more exist and can be found.

Answer 1

I think the answer is:

Ham words are words that consist solely of characters, whose binary representation (ASCII) has even parity, i.e. contains an even number of 1's.
These are called Ham words, because they are related to the Hamming weight of the binary representation of their letters. The Hamming weight of a binary code is the sum of all 1's in it. All Ham words are words in which all letters have even Hamming weight, which we might call Ham letters.
The full list of (uppercase) letters goes:
A = 1000001 --> Hamming weight = 2 --> Ham letter
B = 1000010 --> Hamming weight = 2 --> Ham letter
C = 1000011 --> Hamming weight = 3 --> Not a Ham letter
D = 1000100 --> Hamming weight = 2 --> Ham letter
E = 1000101 --> Hamming weight = 3 --> Not a Ham letter
F = 1000110 --> Hamming weight = 3 --> Not a Ham letter
G = 1000111 --> Hamming weight = 4 --> Ham letter
H = 1001000 --> Hamming weight = 2 --> Ham letter
I = 1001001 --> Hamming weight = 3 --> Not a Ham letter
J = 1001010 --> Hamming weight = 3 --> Not a Ham letter
K = 1001011 --> Hamming weight = 4 --> Ham letter
L = 1001100 --> Hamming weight = 3 --> Not a Ham letter
M = 1001101 --> Hamming weight = 4 --> Ham letter
N = 1001110 --> Hamming weight = 4 --> Ham letter
O = 1001111 --> Hamming weight = 5 --> Not a Ham letter
P = 1010000 --> Hamming weight = 2 --> Ham letter
Q = 1010001 --> Hamming weight = 3 --> Not a Ham letter
R = 1010010 --> Hamming weight = 3 --> Not a Ham letter
S = 1010011 --> Hamming weight = 4 --> Ham letter
T = 1010100 --> Hamming weight = 3 --> Not a Ham letter
U = 1010101 --> Hamming weight = 4 --> Ham letter
V = 1010110 --> Hamming weight = 4 --> Ham letter
W = 1010111 --> Hamming weight = 5 --> Not a Ham letter
X = 1011000 --> Hamming weight = 3 --> Not a Ham letter
Y = 1011001 --> Hamming weight = 4 --> Ham letter
Z = 1011010 --> Hamming weight = 4 --> Ham letter
The lowercase letters are the exact opposite, because all the 0's in the second position will be replaced by 1's, increasing the Hamming weight by 1 and thereby reversing the parity.
In other words:
Ham letters: ABDGHKMNPSUVYZ and cefijloqrtwx
Not Ham letters: CEFIJLOQRTWX and abdghkmnpsuvyz

Reasoning that led to this conclusion:

First of all, there's a lot of similar letters in all of the Ham words, so I'm assuming a word is a Ham word if it consists solely of letters that are within a certain subset.
If this is indeed true, all we need to do here is determine the subset and why it was chosen in that specific way.
First, we can collect all letters that are definitely in the subset (namely the letters we observe in Ham words), they are the following (and I will split this out in uppercase and lowercase letters, since they should clearly be treated separately): BDAMPGUNSHK, ertowlxicf, or, reordering: ABDGHKMNPSU and cefilortwx Next, we can collect all letters that are definitely not in the subset. The way we do this is, we look for Not-Ham words that are made up out of letters from the subset (as mentioned above) and one single letter not from the subset: CR, s. Unfortunately, there aren't many of them.

What is striking about this:

The letters in the uppercase subset are fully disjoint from the letters in the lowercase subset, i.e. none of the letters in the uppercase subset occur in the lowercase subset and vice versa.
This means we can extend our collection of letters definitely not in the subset to: CEFILORTWX and abdghkmnpsu.
Also, the following letters are unknowns: JQVYZ
Now to figure out why these subsets were actually chosen in this way..