MySQL 8.0

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create table example_table as
select 1 as userid, 'DOG' as license union all
select 1,   'CAT' union all
select 1,   'FISH' union all
select 2,   'DOG' union all
select 2,   'CAT' union all
select 3,   'FISH' union all
select 4,   'CAT' union all
select 4,   'DOG' union all
select 4,   'FISH' union all
select 5,   'CAT' union all
select 5,   'FISH';
SELECT *
  FROM example_table


userid
license


1
DOG

1
CAT

1
FISH

2
DOG

2
CAT

3
FISH

4
CAT

4
DOG

4
FISH

5
CAT

5
FISH


SELECT USERID, COUNT(*) AS LICENSE_COUNT
  FROM example_table
  GROUP BY USERID
  HAVING COUNT(*) = 3


USERID
LICENSE_COUNT


1
3

4
3