Let r and s be partial orders on a nonempty set a prove that t = r intersection s is also a partial order on a.

See proof below

Step-by-step explanation:

We denote (x,y)∈R as xRy, and we also use the similar notation xSy for (x,y)∈S. Remember that R and S are reflexive, antisymmetric and transitive relations (the definition of partial order).

To prove that R∩S⊆A is a partial order, we will prove that R∩S is reflexive, antisymmetric and transitive.

Reflexive: Let a∈A. R is reflexive thus aRa. S is also reflexive, then aSa. Then (a,a)∈R and (a,a)∈S which implies that (a,a)∈R∩S, that is, a(R∩S)a for all a∈A.Antisymmetric: Let a,b∈A and suppose that a(R∩S)b and b(R∩S)a hold. In particular, aRb and bRa. Since R is antisymmetric, a=b.Transitive: Let a,b,c∈A and suppose that a(R∩S)b and b(R∩S)c hold. Then aRb, bRc, aSb and bSc are true. The first two statements imply by the transitivity of R that aRc. Similarly, from the last two we have that aSc. Thus a(R∩S)c as we wanted to prove.

2+1=3 that’s your answer