II. We may next take the more complex example contained in the same extract from Jevons.
The given alternants are ACe, aBCe, aBcdE, abCe, abcE ; and by the reduction of dual terms, they become aBcdE, abcE, Ce. Therefore, What is not aBcdE or abcE is Ce ; and this proposition may be resolved into the four propositions:—
| ⎧ ⎨ ⎩ | All A is Ce ; | (1) |
| All BD is Ce ; | (2) | |
| All C is e ; | (3) | |
| All e is C. | (4) |
But since by (3) All C is e, (1) may be reduced to All A is C ; and this proposition may be combined with (4) yielding All c is aE. Also by (3), (2) may be reduced to All BD is C.
Hence our solution becomes
| ⎧ | All BD is C, |
| ⎨ | All C is e, |
| ⎩ | All c is aE. |
This solution may be shewn to be equivalent to the solution given by Jevons himself.
III. The following problem is from Jevons, Principles of Science, 2nd ed., p. 127 (Problem v).
The given alternants are ABCD, ABCd, ABcd, AbCD, AbcD, aBCD, aBcD, aBcd, abCd.
By the reduction of duals these alternants may be written as follows: ABC or ABcd or AbD or aBCD or aBc or abCd.