Considere os comandos SQL abaixo, que foram executados no MySQL 5.5:

create table produto (
  id int not null primary key, 
  nome varchar(50) not null ); 
create table notafiscal (
  num int not null primary key, 
  emissao datetime not null, 
  cliente varchar(100) not null); 
create table notaitem (
  num int not null, ordem int not null, 
  produto int not null, 
  qdade int not null, 
  valor_unit numeric(10,2) not null, 
  primary key (num,ordem), 
  constraint nin foreign key (num) 
    references notafiscal (num) 
      on delete cascade 
      on update cascade constraint nip 
    foreign key (produto) references produto (id) 
      on delete restrict 
      on update cascade );
insert into produto values 
  (1,'A'),(2,'B'),(3,'C'),(4,'D'),(5,'E'),(6,'F'),(7,'G'),(8,'H'); 
insert into notafiscal values 
  (1,'2018-01-10','cliente 1'),
  (2,'2018-01-10','cliente 2'),
  (3,'2018-01-12','cliente 1'), 
  (4,'2018-01-13','cliente 3'),
  (5,'2018-01-13','cliente 5'),  
  (6,'2018-01-13','cliente 1'),
  (7,'2018-01-14','cliente 2'),
  (8,'2018-01-14','cliente 4'),
  (9,'2018-01-15','cliente 5'),
  (10,'2018-01-16','cliente 6'),
  (11,'2018-01-17','cliente 7');
insert into notaitem values 
  (1,1,1,1,10.0),
  (1,2,3,1,15.0), 
  (2,1,1,1,10.0),
  (2,2,3,1,15.0),
  (2,3,5,1,5.0), 
  (3,1,1,1,10.0),
  (3,2,6,1,20.0), 
  (4,1,1,1,10.0),
  (4,2,3,7,5.0),
  (4,3,2,1,15.0),
  (4,4,6,1,20.0), 
  (5,1,1,1,10.0),
  (5,2,3,1,15.0), 
  (7,1,2,1,15.0),
  (7,2,4,1,10.0);

Assinale a alternativa que apresenta a instrução do SQL que fornece os resultados apresentados na tabela abaixo.