[HACKERS] Constraint exclusion failed to prune partition in case of partitionexpression involves function call

Hi,

In following case, constraint exclusion not able prune partition (even
if function is immutable), is this know behaviour?

--CASE 1 :  create table & insert data
create table foo_list (a integer, b text) partition by list (abs(a));
create table foo_list1 partition of foo_list for values in (0);
create table foo_list2 partition of foo_list for values in (1);
create table foo_list3 partition of foo_list for values in (2);
create table foo_list4 partition of foo_list for values in (3);
insert into foo_list values(0),(1),(-1),(2),(-2),(3),(-3);

--Explain plan
postgres=# explain select * from foo_list where a = 2;                          QUERY PLAN
-----------------------------------------------------------------Append  (cost=0.00..103.50 rows=25 width=36)  ->  Seq
Scanon foo_list  (cost=0.00..0.00 rows=1 width=36)        Filter: (a = 2)  ->  Seq Scan on foo_list1  (cost=0.00..25.88
rows=6width=36)        Filter: (a = 2)  ->  Seq Scan on foo_list2  (cost=0.00..25.88 rows=6 width=36)        Filter: (a
=2)  ->  Seq Scan on foo_list3  (cost=0.00..25.88 rows=6 width=36)        Filter: (a = 2)  ->  Seq Scan on foo_list4
(cost=0.00..25.88rows=6 width=36)        Filter: (a = 2)
 
(11 rows)

AFAUI, constraint exclusion should prune all above table other than
foo_list3 as happens in the following case :

-- CASE 2: create table & insert data
create table bar_list (a integer, b text) partition by list (a);
create table bar_list1 partition of bar_list for values in (0);
create table bar_list2 partition of bar_list for values in (1);
create table bar_list3 partition of bar_list for values in (2);
create table bar_list4 partition of bar_list for values in (3);
insert into bar_list values(0),(1),(2),(3);

--- Explain plan
postgres=# explain select * from bar_list where a = 2;                          QUERY PLAN
-----------------------------------------------------------------Append  (cost=0.00..25.88 rows=7 width=36)  ->  Seq
Scanon bar_list  (cost=0.00..0.00 rows=1 width=36)        Filter: (a = 2)  ->  Seq Scan on bar_list3  (cost=0.00..25.88
rows=6width=36)        Filter: (a = 2)
 
(5 rows)

Thanks & Regards,
Amul