sharkiefff/RBAC-Text2SQL-Benchmark huggingface dataset:
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: What are the names of the different customers who have taken out a loan, ordered by the total amount that they have taken?
Answer: SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id  =  T2.cust_id GROUP BY T1.cust_name ORDER BY sum(T2.amount)
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id  =  T2.cust_id GROUP BY T1.cust_name ORDER BY sum(T2.amount)
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: Find the state, account type, and credit score of the customer whose number of loan is 0.
Answer: SELECT state ,  acc_type ,  credit_score FROM customer WHERE no_of_loans  =  0
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT state ,  acc_type ,  credit_score FROM customer WHERE no_of_loans  =  0
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: What are the states, account types, and credit scores for customers who have 0 loans?
Answer: SELECT state ,  acc_type ,  credit_score FROM customer WHERE no_of_loans  =  0
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT state ,  acc_type ,  credit_score FROM customer WHERE no_of_loans  =  0
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: Find the number of different cities which banks are located at.
Answer: SELECT count(DISTINCT city) FROM bank
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT count(DISTINCT city) FROM bank
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: In how many different cities are banks located?
Answer: SELECT count(DISTINCT city) FROM bank
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT count(DISTINCT city) FROM bank
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: Find the number of different states which banks are located at.
Answer: SELECT count(DISTINCT state) FROM bank
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT count(DISTINCT state) FROM bank
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: In how many different states are banks located?
Answer: SELECT count(DISTINCT state) FROM bank
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT count(DISTINCT state) FROM bank
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: How many distinct types of accounts are there?
Answer: SELECT count(DISTINCT acc_type) FROM customer
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT count(DISTINCT acc_type) FROM customer
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: Count the number of different account types.
Answer: SELECT count(DISTINCT acc_type) FROM customer
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT count(DISTINCT acc_type) FROM customer
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: Find the name and account balance of the customer whose name includes the letter ‘a’.
Answer: SELECT cust_name ,  acc_bal FROM customer WHERE cust_name LIKE '%a%'
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT cust_name ,  acc_bal FROM customer WHERE cust_name LIKE '%a%'
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: What are the names and account balances of customers with the letter a in their names?
Answer: SELECT cust_name ,  acc_bal FROM customer WHERE cust_name LIKE '%a%'
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT cust_name ,  acc_bal FROM customer WHERE cust_name LIKE '%a%'
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: Find the total account balance of each customer from Utah or Texas.
Answer: SELECT sum(acc_bal) FROM customer WHERE state  =  'Utah' OR state  =  'Texas'
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT sum(acc_bal) FROM customer WHERE state  =  'Utah' OR state  =  'Texas'
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: What are the total account balances for each customer from Utah or Texas?
Answer: SELECT sum(acc_bal) FROM customer WHERE state  =  'Utah' OR state  =  'Texas'
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT sum(acc_bal) FROM customer WHERE state  =  'Utah' OR state  =  'Texas'
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: Find the name of customers who have both saving and checking account types.
Answer: SELECT cust_name FROM customer WHERE acc_type  =  'saving' INTERSECT SELECT cust_name FROM customer WHERE acc_type  =  'checking'
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT cust_name FROM customer WHERE acc_type  =  'saving' INTERSECT SELECT cust_name FROM customer WHERE acc_type  =  'checking'
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: What are the names of customers who have both savings and checking accounts?
Answer: SELECT cust_name FROM customer WHERE acc_type  =  'saving' INTERSECT SELECT cust_name FROM customer WHERE acc_type  =  'checking'
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT cust_name FROM customer WHERE acc_type  =  'saving' INTERSECT SELECT cust_name FROM customer WHERE acc_type  =  'checking'
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: Find the name of customers who do not have an saving account.
Answer: SELECT cust_name FROM customer EXCEPT SELECT cust_name FROM customer WHERE acc_type  =  'saving'
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT cust_name FROM customer EXCEPT SELECT cust_name FROM customer WHERE acc_type  =  'saving'
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: What are the names of customers who do not have saving accounts?
Answer: SELECT cust_name FROM customer EXCEPT SELECT cust_name FROM customer WHERE acc_type  =  'saving'
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT cust_name FROM customer EXCEPT SELECT cust_name FROM customer WHERE acc_type  =  'saving'
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: Find the name of customers who do not have a loan with a type of Mortgages.
Answer: SELECT cust_name FROM customer EXCEPT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id  =  T2.cust_id WHERE T2.loan_type  =  'Mortgages'
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT cust_name FROM customer EXCEPT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id  =  T2.cust_id WHERE T2.loan_type  =  'Mortgages'
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: What are the names of customers who have not taken a Mortage loan?
Answer: SELECT cust_name FROM customer EXCEPT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id  =  T2.cust_id WHERE T2.loan_type  =  'Mortgages'
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT cust_name FROM customer EXCEPT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id  =  T2.cust_id WHERE T2.loan_type  =  'Mortgages'
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: Find the name of customers who have loans of both Mortgages and Auto.
Answer: SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id  =  T2.cust_id WHERE loan_type  =  'Mortgages' INTERSECT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id  =  T2.cust_id WHERE loan_type  =  'Auto'
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id  =  T2.cust_id WHERE loan_type  =  'Mortgages' INTERSECT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id  =  T2.cust_id WHERE loan_type  =  'Auto'
##Instruction:
Database: loan_1
Total Tables: 3
Total Columns: 18

Table Schemas:

loan_1 contains tables such as bank, customer, loan.
Table bank has columns such as branch_ID, bname, no_of_customers, city, state.
branch_ID is the primary key.
Table customer has columns such as cust_ID, cust_name, acc_type, acc_bal, no_of_loans, credit_score, branch_ID, state.
cust_ID is the primary key.
Table loan has columns such as loan_ID, loan_type, cust_ID, branch_ID, amount.
loan_ID is the primary key.
The branch_ID of customer is the foreign key of branch_ID of bank.
The branch_ID of loan is the foreign key of branch_ID of bank.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns:
  - bank: bname, branch_ID, city, state
  - customer: * (all columns)
  - loan: * (all columns)

Question: What are the names of customers who have taken both Mortgage and Auto loans?
Answer: SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id  =  T2.cust_id WHERE loan_type  =  'Mortgages' INTERSECT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id  =  T2.cust_id WHERE loan_type  =  'Auto'
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id  =  T2.cust_id WHERE loan_type  =  'Mortgages' INTERSECT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id  =  T2.cust_id WHERE loan_type  =  'Auto'