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

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

 

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