Some Remarks on Relational Assignment - View Updating and Relational Theory - page 232

reference R in the expression on the right side of the second assignment denotes the value of
relvar R after the first assignment has been executed):

R := R - d1 i1 , R := R - d2 ∪ ∪ i2 ;

Now, the following pairs of sets aren't necessarily disjoint, in general: d1 and d2 ; i1 and i2 ; d1
and i2 ; d2 and i1 . However, it's fairly easy to see (exercise for the reader!) that the foregoing
double assignment is logically equivalent to a single assignment of the form—

R := r - d

∪ i ;

—where:

 r is the initial value of R (i.e., the value of R before the first assignment is done)

 d = ( d1 - i2 )

d2

∪

 i = ( i1 ∪ i2 ) - d2

Note that d and i here are certainly disjoint; however, it's not the case, in general, that d is
included in r or that i and r are disjoint. Now define:

 d′ = ( r ∩ d ) - i

 i′ = i - ( r ∪ d )

Then the original double assignment is logically equivalent to the following single assignment—

R := r - d′ ∪ i′ ;

—where d′ - r = r i′ = Ø.

Note finally that it follows by induction from all of the above that a multiple assignment of
the form

i′ = d′

∩ ∩

R := rx1 , R := rx2 , ... , R := rxn
for arbitrary n > 0 can always be reduced to a single assignment of the form

R := r - d

∪ i

where r is the initial value of R and d - r = r i = d i = Ø.

∩ ∩