Suppose R is a partial order on A and B ? A. Prove that R ?(BxB) is a partial or

Question

Suppose R is a partial order on A and B?A. Prove that R?(BxB) is a partial order on B.

Can anyone edit my attempt below or fix it please if needed?
My attempt:

* If R is a partial order, then R is reflexive, antisymmetric, and transitive.

* If R is reflexive, then if a?A, aRa

* If R is antisymmetric, then if a,b?A, then [aRb ^ bRa] --> a=b

* If R is transitive, then if a,b,c?A, [aRb ^ bRc] --> aRc

* B?A means if (x,y)?B, then (x,y)?A right?

* R?(BxB) means (a, (x,y)) ? R?(BxB) iff a?R ^ (x,y)?(BxB) iff a?R & x?B & y?B correct?

* R?(BxB) is a partial order on B means that if x,y,z?B, then R?(BxB) is reflexive, antisymmetric, and transitive right?

* If R?(BxB) is reflexive, then if x?B, then xRx & if a?A, then aRa
* if R?(BxB) is antisymmetric, then if x,y?B, then [xRy & yRx] --> x=y, and if a,b?A, then [aRb ^ bRa] --> a=b

* if R?(BxB) is transitive, then if x,y,z?B, [xRy & yRz] --> xRz, and if a,b,c?A, [aRb ^ bRc] --> aRc

* Now since B?A & x,y,z?B, then x,y,z?A
* Since R is a relation on A & R is a partial order, then xRx, [xRy & yRx] --> x=y, & [xRy & yRz] --> xRz
* Since R is a partial order on A, if a,b,c?A, aRa, [aRb ^ bRa] --> a=b, and [aRb ^ bRc] --> aRc
Therefore
R?(BxB) is a partial order on B

Explanation / Answer

Let x be an arbitrary element of B. Then $(x,x) in B X B$, also $(x,x) in A X A$, so $(x,x) in R$. Thus $(x,x) in R cap (B X B)$. Since x is arbitrary, so $R cap (B X B)$ is reflexive on B

Let x,y,z be arbitrary elements of B s.t. $(x,y) in R cap (B X B)$ and $(y,z) in R cap (B X B)$. It follows that $(x,y) in R$ and $(y,z) in R$. Since R is a partial order on A, so R is transitive and hence $(x,z) in R$. Also $(x,z) in B X B$. Thus $(x,z) in R cap (B X B)$. Since x,y,z are arbitrary so $R cap (B X B)$ is transitive on B

Let x,y be arbitrary elements of B s.t. $(x,y) in R cap (B X B)$. Then $(x,y) in R$. Since R is partial order and hence antisymmetric on A, so $(x,y) in R land (y,x) in R ightarrow x = y$. So $(x,y) in (R cap (B X B)) land (y,x) in (R cap (B X B)) ightarrow x = y$. Since x,y are arbitrary so $R cap (B X B)$ is antisymmetric on B

Thus $R cap (B X B)$ is partial order on B.

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