{[], se_lt_x :: [a-A, c-C] =*= (se_lt_y :: [d-A] ; se_lt_z :: [g-A])}
The generated SQL is:
SELECT x43.c FROM (se_lt_x x43 INNER JOIN se_lt_z z6 ON x43.a=z6.g AND z6.g=x43.a) UNION SELECT x44.c FROM (se_lt_x x44 INNER JOIN se_lt_y y16 ON x44.a=y16.d AND y16.d=x44.a)