This is a semi-ordered relationship, but it is not a fully ordered relationship.
Verification is basically trivial, and the corresponding properties of R can be obtained from reflexivity, antisymmetry and transitivity ≤.
It's simple if it's not a complete sequence. If a ≠ b, then
Otherwise, a ≤ b∧b ≤ a, and a = b is antisymmetric from ≤, which is contradictory.
2. The binding relationship is (x≤u∧x≠u)∩(x = u∧y≤v), right?
This is the dictionary order, which is a full-order relationship and therefore a semi-order relationship. A×A is a finite set and a well-ordered relationship.
Anti-symmetry: If
By < x, y >R<u, v> is (x≤u∧x≠u)∩(x = u∧y≤v),
Get (x ≤ u∧x ≠ u)∨x = u, that is, x ≤ u.
The same reason
So x = u is obtained from the antisymmetry of ≤
Replace < x, y >R<u, v> to get y ≤ v, substitute < u, v>R<x, y> to get v ≤ y.
Then y = v is obtained from the antisymmetry of ≤, so < x, y >; = & ltu，v & gt。
Transitivity: If
By < x, y >R<u, v>X ≤ u, defined as
If x ≠ s, then
If x = s and u ≤ s = x, u = x (≤ antisymmetry) can be obtained, so x = u = s.
Replace < x, y >R<u, v> to get y ≤ v, substitute < u, v>R<s, t> to get v ≤ T. So from ≤ transitivity to get y ≤ T.
Know that < x, y>R<s, t> is also established.
Integrity: any
From the completeness of ≤, x ≤ u or u ≤ x holds. Suppose x ≤ u.
If x ≠ u, there is
If x = u, when y ≤ v, there is
But from the completeness of ≤, at least one of y ≤ v or v ≤ y holds.
therefore
3. It is not a semi-ordered relation, because there is no antisymmetry.
For a ≠ b, the completeness from ≤ can be set as a ≤ b. known
4. It is not a semi-ordered relationship, because there is no reflexivity. That is to say,
Personally, I am not very familiar with the language of discrete mathematics. If you have any questions, ask me.