A COMPLETE AND CONCRETE EXAMPLE
The reader may still be wondering when he will see a complete and concrete example of the application of the logical-truth calculator. So far we have given only pieces of examples in order to illustrate some explanation. Therefore, let us consider now the following problem:
Problem. The A. A. Adams Co., Inc., has about 1000 employees. About 600 of them are insured under a contract for group insurance with the I. I. Insurance Co. Mr. Adams decides that more of his employees ought to be insured. As a part of his study of the change, he asks his manager in charge of the group insurance plan, “What are the possible statuses of my employees who are not insured?”
The manager replies, “I can tell you the names of the men who are not insured, and all the data you may want to know about them.”
Mr. Adams says, “No, John, that won’t be enough, for I need to know whether there are any groups or classes that for some basic reason I should exclude from the change I am considering.”
So the manager goes to work with the following 5 statuses and the following 5 rules, and he produces the following answer. Our question is, “Is he right, or has he made a mistake?”
Statuses. A status for any employee is a report about that employee, answering all the following 5 questions with “yes” or “no.”
1. Is the employee eligible for insurance?
2. Has the employee applied for insurance?
3. Has the employee’s application for insurance been approved?
4. Does the employee require a medical examination for insurance?
5. Is the employee insured?
Rules. The rules applying to employees are:
A. Any employee, to be insured, must be eligible for insurance, must make application for insurance, and must have such application for insurance approved.
B. Only eligible employees may apply for insurance.
C. The application of any person eligible for insurance without medical examination is automatically approved.
D. (Naturally) an application can be approved only if the application is made.
E. (Naturally) a medical examination will not be required from any person not eligible for insurance.
Answer by the Manager. There are 5 possible combinations of statuses for employees who are not insured, as shown in [Table 15].
Table 15
| Possible Combination of Statuses | Status 1, Eligible | Status 2, Applied | Status 3, Application Approved | Status 4, Examination Required | Status 5, Insured |
|---|---|---|---|---|---|
| 1 | Yes | Yes | Yes | Yes | No |
| 2 | Yes | Yes | Yes | No | No |
| 3 | Yes | Yes | No | Yes | No |
| 4 | Yes | No | Yes | No | No |
| 5 | No | No | No | No | No |
The question may be asked why employees who are eligible, who have applied for insurance, who have had their applications approved, and who require no medical examination (combination 2) are yet not insured. The answer is that the rules given do not logically lead to this conclusion. As a matter of fact, there might be additional rules, such as: any sick employee must first return to work; or any period from date of approval of application to the first of the following month must first pass.
The first step in putting this problem on the Kalin-Burkhart Logical-Truth Calculator is to rephrase the rules, using the language of the connectives that we have on the machine. The rules rephrased are:
A. If an employee is insured, then he is eligible, he has applied for insurance, and his application has been approved.
if 5, then 1 and 2 and 3
B. If an employee has applied (under these rules) for insurance, then he is eligible.
if 2, then 1
C. If an employee is eligible for insurance, has applied, and requires no medical examination, his application is automatically approved.
if 1 and 2 and not-4, then 3
D. If an employee’s application has been approved, then he has applied.
if 3, then 2
E. If an employee is not eligible, then he does not require a medical examination (under these rules).
if not-1, then not-4
To get the answer we seek, we must add one more rule for this answer only:
F. The employee is not insured.
not-5
We now have a total of 4 + 2 + 4 + 2 + 2 + 1 occurrences of statements, or 15 occurrences. This is beyond the capacity of the existing machine. But fortunately Rule F and Rule A cancel each other; they may both be omitted; and this gives us 10 occurrences instead of 15. In other words, all the possible statuses under “Rule B and Rule C and Rule D and Rule E” will give us the answer we seek.
The rephrasing and reasoning we have done here is perhaps not easy. For example, going from the logical pattern
Only igs may be ows
to the logical pattern
If it is an ow, then it is an ig
as we did in rephrasing Rule B, deserves rather more thought and discussion than we can give to the subject here. A person who is responsible for preparing problems for the Logical-Truth Calculator should know the algebra of logic.
Choosing an appropriate grouping, we now set on the machine:
{(if 2, then 1) and [IF (1 and 2) and not-4, then 3]} and
[(if 3, then 2) and (IF not-1, then not-4)]
The setting is as shown in [Table 16]. After this setting, the machine is turned on and set to stop on the “true” cases. The
Table 16
SETTING OF THE PROBLEM
ON THE
LOGICAL-TRUTH CALCULATOR
| Unit | |||||||||||||
| 3 | Statement Dial No. | V₁ | V₂ | V₃ | V₄ | V₅ | V₆ | V₇ | V₈ | V₉ | V₁₀ | V₁₁ | V₁₂ |
| 3 | Statement Dial Setting | 2 | 1 | 1 | 2 | 4 | 3 | 3 | 2 | 1 | 4 | F | F |
| 2 | Statement Denial Switch | ||||||||||||
| Setting | — | — | — | — | up | — | — | — | up | up | — | — | |
| 4 | Connective Dial No. | k₁ | k₂ | k₃ | k₄ | k₅ | k₆ | k₇ | k₈ | k₉ | k₁₀ | k₁₁ | |
| 4 | Connective Dial Setting | ▲ | ∧ | ∧ | ∧ | ▲ | ∧ | ▲ | ∧ | ▲ | off | off | |
| 8 | Statement Denial Switch | ||||||||||||
| Setting | — | — | — | — | — | — | — | — | — | — | — | ||
| 5 | Antecedent Dial No. | A₁ | A₂ | A₃ | A₄ | A₅ | A₆ | A₇ | A₈ | A₉ | A₁₀ | A₁₁ | |
| 5 | Antecedent Dial Setting | V | k₁ | V | k₃ | k₄ | k₂ | V | k₇ | V | off | off | |
| 6 | Consequent Dial No. | C₁ | C₂ | C₃ | C₄ | C₅ | C₆ | C₇ | C₈ | C₉ | C₁₀ | C₁₁ | |
| 6 | Consequent Dial Setting | V | k₅ | V | V | V | k₈ | V | k₉ | V | off | off | |
| 7 | Stop Switches, associating | ||||||||||||
| connective to Main | |||||||||||||
| Truth-Value Light | — | — | — | — | — | — | — | — | — | — | — | ||
possible statuses of employees who are not insured are shown in [Table 17]. As we look down the last column in [Table 17], we observe 6 occurrences of T, instead of 5 as the manager determined ([see Table 15]). Thus, when we compare the manager’s result with the machine result, we find an additional possible combination to be reported to Mr. Adams, combination 7:
Employee eligible, employee has not applied, employee’s application not approved, employee requires a medical examination, employee not insured.
Table 17
SOLUTION OF THE PROBLEM
BY THE CALCULATOR
- LEGEND:
- {A} THE EMPLOYEE IS ELIGIBLE FOR INSURANCE
- {B} THE EMPLOYEE HAS APPLIED FOR INSURANCE
- {C} THE EMPLOYEE’S APPLICATION FOR INSURANCE
- HAS BEEN APPROVED
- {D} THE EMPLOYEE REQUIRES A MEDICAL EXAMINATION
- {E} THE EMPLOYEE IS INSURED
- {F} CASE, OR COMBINATION NO.
- {G} THE COMBINATION DOES NOT CONTRADICT THE RULES,
- I.E., THE YELLOW LIGHT IS ON
| {A} | {B} | {C} | {D} | {E} | {F} | {G} | ||
|---|---|---|---|---|---|---|---|---|
| Status: | 1 | 2 | 3 | 4 | 5 | |||
| T | T | T | T | F | 1 | T | ||
| F | T | T | T | F | 2 | F | ||
| T | F | T | T | F | 3 | F | ||
| F | F | T | T | F | 4 | F | ||
| T | T | F | T | F | 5 | T | ||
| F | T | F | T | F | 6 | F | ||
| T | F | F | T | F | 7 | T | ||
| F | F | F | T | F | 8 | F | ||
| T | T | T | F | F | 9 | T | ||
| F | T | T | F | F | 10 | F | ||
| T | F | T | F | F | 11 | F | ||
| F | F | T | F | F | 12 | F | ||
| T | T | F | F | F | 13 | F | ||
| F | T | F | F | F | 14 | F | ||
| T | F | F | F | F | 15 | T | ||
| F | F | F | F | F | 16 | T |
Because of the medical examination, this additional class of employee would need to be considered rather carefully in any change of the group insurance plan.