user_language.sql

create table stackoverflow_user (
  user_name varchar(100) primary key
);

create table stackoverflow_language (
  language_name varchar(100) primary key
);

create table stackoverflow_user_language (
  user_name varchar(100) references stackoverflow_user(user_name),
  language_name varchar(100) references stackoverflow_language(language_name),
  primary key (user_name, language_name)
);

insert into stackoverflow_user values ('bob'), ('alice'), ('mallory'), ('robert');
insert into stackoverflow_language (language_name) values ('english'), ('spanish'), ('italian'), ('japanese');

insert into stackoverflow_user_language (user_name, language_name) values
  ('robert', 'spanish'), ('robert', 'italian')
, ('bob', 'spanish')
, ('mallory', 'italian')
, ('alice', 'japanese')
;

select u.user_name, count(l.language_name)
  from stackoverflow_user u
  join stackoverflow_user_language ul on u.user_name = ul.user_name
  join stackoverflow_language l on l.language_name = ul.language_name
 where l.language_name in ('spanish', 'italian')
 group by u.user_name
 having count(l.language_name) = 2