SQL statement to prove that a candidate key constraint holds in a R(ABCD)

Question

How do I write a SQL statement that proves the candidate key ACD holds given a relation with attributes ABCD and the functional dependency A → B ? I know there's something similar here: SQL statement to prove that A->B in a R(ABCD), but can't figure out how to write the query for this constraint.

Answer 1

\i tmp.sql
create table abcd(
        a integer not null
        ,b integer not null
        ,c integer not null
        ,d integer not null
        -- , PRIMARY KEY (a,b,c,d)
        );
INSERT INTO abcd(a,b,c,d)
select (s/4)%4, (s/4)%2,(s/2)%2,s%2
from generate_series(0,15) s
        ;
select *from abcd;

ALTER TABLE abcd ADD UNIQUE (a,b,c,d); --succeeds
ALTER TABLE abcd ADD UNIQUE (a,c,d); --succeeds
ALTER TABLE abcd ADD UNIQUE (b,c,d); --fails

---

Results:

---

DROP SCHEMA
CREATE SCHEMA
SET
CREATE TABLE
INSERT 0 16
 a | b | c | d 
---+---+---+---
 0 | 0 | 0 | 0
 0 | 0 | 0 | 1
 0 | 0 | 1 | 0
 0 | 0 | 1 | 1
 1 | 1 | 0 | 0
 1 | 1 | 0 | 1
 1 | 1 | 1 | 0
 1 | 1 | 1 | 1
 2 | 0 | 0 | 0
 2 | 0 | 0 | 1
 2 | 0 | 1 | 0
 2 | 0 | 1 | 1
 3 | 1 | 0 | 0
 3 | 1 | 0 | 1
 3 | 1 | 1 | 0
 3 | 1 | 1 | 1
(16 rows)

ALTER TABLE
ALTER TABLE
ERROR:  could not create unique index "abcd_b_c_d_key"
DETAIL:  Key (b, c, d)=(0, 0, 0) is duplicated.

---

B is functionally dependent on A here, but multiple As can point to the same B value.

BTW: IMO it is impossible to prove something in SQL (it depends in the current data in the table(s)), but it may be possible to reject it. (by composing an example)