Let $a, b, c \in \mathbf{N}_1$ and triples $(a, b, c)$ are formed from them randomly, for example, $(1, 5, 6)$, $(1, 2, 3)$, $(1, 2, 4)$. The order of numbers does not matter in a triplet. We can use the two binary relations $"<"$ and $"="$ and the sum operator $"+"$.
I need to find all possible relations for numbers $a$, $b$, $c$.
I have found the next cases:
1) numbers are the same,
2) numbers are different,
3) numbers are different and sum of first and second numbers equals to the third one,
4) two numbers are the same and their sum is less than the third number,
5) two numbers are the same and their sum is greate than the third number.
My attempt
Edit after the Roddy MacPhee's answer.
We have four base cases: $$a=b=c, \tag{1}$$ $$a<b<c, \tag{2}$$ $$a=b<c, \tag{3}$$ $$a<b=c. \tag{4}$$ The relation (1) is trivial one. Recall that the order of numbers $a$, $b$, $c$ does not matter in a triplet $(a,b,c)$. Therefore, we can omit the relation (4) and use the relation (3) only. The relations (2), (3) have 3 cases each: $$ c<a+b, \quad c=a+b, \quad a+b<c, \tag{2.1}$$ $$ c<a+a, \quad c=a+a, \quad a+a<c. \tag{3.1}$$ Finally, we have 7 cases: $$a=b=c, \tag{A}$$ $$a<b<c \quad \text{and} \quad c < a+b \tag{B}$$ $$a<b<c \quad \text{and} \quad c=a+b \tag{C}$$ $$a<b<c \quad \text{and} \quad a+b<c \tag{D}$$ $$a=b<c \quad \text{and} \quad c<a+a \tag{E}$$ $$a=b<c \quad \text{and} \quad c=a+a \tag{F}$$ $$a=b<c \quad \text{and} \quad a+a<c \tag{G}$$
Question. Were all cases considered with the relation set (<, =) and the sum operator '+'?
We have four base cases: $$a<b<c, \tag{1}$$ $$a=b<c, \tag{2}$$ $$a=b=c, \tag{3}$$ $$a<b=c. \tag{4}$$
The relations (1), (2) have 3 cases each: $$c<a+b, \tag{1.1}$$ $$c=a+b, \tag{1.2}$$ $$a+b<c, \tag{1.3}$$
As does the relation (4): $$b<a+a, \tag{4.1}$$ $$ b=a+a, \tag{4.2}$$ $$ a+a<b. \tag{4.3}$$
Which also works, for the first part, of the first one.
This shows the first one has 9 cases, the second has at least 3 cases, the third has 1 case, and the last has 3 cases. All told that means we have at least 16 cases (assuming I haven't missed things, or messed up, exactly 16).
To sum up, no. You definitely had nowhere near enough cases( at least if you want all possible comparisons).
EDIT
Realized, I can do three cases of b and c as well if not equal. which makes it even more off the mark. All these 3 case ones are a version of the Trichotomy law .