I try to understand how the relational division can be expressed by primitive operations of the relational algebra. So, I found this formula (I writed it such as I found it in the same format):
But I can't get this. If we use $\TIMES$ (the Cartesian product) then A and B relations can't have same attributes, so we need to rename these attributes, i.e.:
If we have $A \{A_1, A_2, A_3\};B \{A_1, A_2, A_4\}$
then the header of the result relation will be
it's okay... But next I need to apply $\MINUS$ but I can't make it because my relation header diffs from $A$.
It means that it requires yet operations (i.e. $\PROJECT$ and $\RENAME$), it isn't? Can anyone explain me more details by this expression?

  • 2
    $\begingroup$ Your pseudo-LaTeX doesn't work. Please fix. What I can read looks wrong. The LHS has no $X$ whereas it is a free variable of the RHS. Finally, it might help if you named the source of the formula that you find confusing. $\endgroup$
    – Kai
    Feb 13, 2017 at 10:50
  • $\begingroup$ @Kai, it's from the book: C.J. Date - An Introduction to Database Systems (8th Edition). $\endgroup$ May 8, 2017 at 0:07

1 Answer 1


The usual form of Division requires that the attributes in B are a subset of the attributes in A. And the result of the Division has just the attributes particular to A -- that is A[X] in your formulation.

I'm not sure if you understand that A[X] is a projection. So the Heading of A[X] TIMES B is exactly the Heading of A. So the inner MINUS is dealing with the same Headings -- that is, Heading of A. Then the [X] at the end of the formula is again a projection; so the outer MINUS is also dealing with the same Headings -- this time [X].

This answers much the same question (using a different notation): Expressing division in relational algebra in terms of other operations See my comments about the various different forms of 'Relational Division'.

And please note: relations have Headings, not 'Headers'.


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