employees of a manufacturing plant/I> and (for
purposes of insurance, etc.) the numbers are noted that are "wedded", "male", "skilled". The Venn
diagram for this is:
PLANT
______________________________________________________
| |
| ____________________ | |
| | | |
| | WEDDED | |
| | ________|____________________ |
| | 1 | | | |
| | | 1 | MALE | |
| | | | | |
| | | | 3 | |
| | _______|_______|________________ | |
| | | 2 | 1 | | | |
| |____|______|_______| | | |
| | | 4 | | |
| | | | | |
| | |_______________________|____| |
| | | |
| | | |
| | 3 | |
| | SKILLED | 5 |
| |______________________________| |
| |
|_____________________________________________________|
Conversely, given data about distribution in sets, the consistency of the report can be checked
by the inclusion-exclusion algorithm. Given: TOTAL (in PLANT) = 20; N(wedded) = 5; N(male)
= 9; N(skilled) = 10; N(wedded & male) = 2; N(wedded & skilled) = 3; N(male & skilled) = 5;
N(wedded & male & skilled) = 1; N(single & female & unskilled) = 5. Then we have the formula:
N(wedded or male or skilled) = N(wedded) + N(male) + N(skilled) - N(wedded & male) - N(wedded & skilled)
- N(male & skilled) + N(wedded & male & skilled) = 5 + 9 + 10 - 2 - 3 - 5 + 1 = 15; N(Total) = N(wedded
or male or skilled) + N(sing & female & unskilled) = 15 + 5 = 20. The report is consistent.
(Note: This would not prove the report to be true. However, an inconsistent report cannot be
true. As Whitehead observed, we can only be sure of what is not so. In a fifty year period, I
have discovered many inconsistent reports issued by Government agencies, coroporation offices, etc. In
one college newspaper -- when it was found that the sum distributed in various categories was fewer than
those reported at the college -- the question was asked: "Are THEY here also?")