Show that functional dependency pseudotransivity holds using closure test
The definition of Pseudotransitivity given above is equivalent to say the following.

From:

(1) A1 A2 ... An -&gt; B1 B2 ... Bm
(2) B1 B2 ... Bm E1 E2 ... Ej -&gt; D


we can derive:

(3) A1 A2 ... An E1 E2 ... Ej -&gt; D


Here is a possible simple proof.

Let’s try to compute the closure of {A1 A2 ... An E1 E2 ... Ej} from the first two functional dependencies. If this closure contains D then we have proved the hypothesis.

{A1 A2 ... An E1 E2 ... Ej}+ = A1 A2 ... An E1 E2 ... Ej    (starting point)
{A1 A2 ... An E1 E2 ... Ej}+ = A1 A2 ... An B1 B2 ... Bm E1 E2 ... Ej (by using 1)
{A1 A2 ... An E1 E2 ... Ej}+ = A1 A2 ... An B1 B2 ... Bm D E1 E2 ... Ej (by using 2)
No other derivation is possible.


So, since D is included in the closure of {A1 A2 ... An E1 E2 ... Ej}, we have shown that A1 A2 ... An E1 E2 ... Ej -&gt; D.