All BC is AD or ad,(ii)
All ab is cd,(iii)
All cd is ab.(iv)
513 By (i), No AB is CD, therefore, No A is BCD. (1)
By (ii), No BC is Ad, therefore, No A is BCd. (2)
Combining (1) and (2), it follows immediately that No A is BC.
Boole also shews that All bC is A. This is a partial contrapositive of (iii). We have so far not required to make use of (iv) at all.
496. Given the same premisses as in the preceding section, prove that:—
(1) Wherever the property C is found, either the property A or the property B will be found with it, but not both of them together;
(2) If the property B is absent, either A and C will be jointly present, or C will be absent;
(3) If A and C are jointly present, B will be absent.[Boole, Laws of Thought, p. 129.]
First, By (i), All C is a or b or d ; by (ii), All C is a or b or D ; therefore, All C is a or b.
Also, by (iii), All C is A or B ;
therefore, All C is Ab or aB. (1)
Secondly, By (iii). All b is A or c,
therefore, by section [432], All b is AC or c. (2)
Thirdly, from (1) it follows immediately that
All AC is b. (3)