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Sheet 8 -Solution Let D 1 ; : : : ; D n be domains. The domain
 

Summary: Sheet 8 - Solution
Exercise 3
Let D 1 ; : : : ; D n be domains. The domain
constructors
and + are de ned as follows:
 D
1
: :
:
D n = f(d 1 ; : : : ; d n ) j d 1 2 D 1 f?D1 g; : : : ; d n 2 D n f?Dn gg[f(?D1 ; : : : ; ?Dn )g
The relation vD
1
:::
Dn is de ned in the same way as vD 1 :::Dn .
 D 1 + : : : +D n = fd D 1 j d 2 D 1 g [ : : : [ fd Dn j d 2 D n g [ f?D 1 +:::+Dn g
The relation vD1+:::+Dn is de ned by e vD1+:::+Dn e 0 i e = ?D1+:::+Dn or e = d D i ,
e 0 = d 0D i and d vD i d 0 for some i 2 f1; : : : ; ng.
Prove or disprove (with consideration of v, but without consideration of the labels):
(i) D 1 +D 2 = D 1? D 2?
(ii) D 1 D 2 = D

  

Source: Ábrahám, Erika - Fachgruppe Informatik, Rheinisch Westfälische Technische Hochschule Aachen (RWTH)

 

Collections: Computer Technologies and Information Sciences