106 
Proceedings of the Royal Society 
circular line ; those having both x and y by the part inside both 
lines, those having neither by the part outside. 
Now xy = xy . \ y = — . 
X 
Expand — in terms' of the parts formed by means of xy and x. 
X 
Xy 
— = ctxyx + b xy (1 — x) + c(l — xy)x-\-d( 1 - xy)( 1 — x) . 
Let xy = 1 and a;= 1. Then, a = = 1 . 
xy — 1 and a? = 0, then h = ^ • 
xy — 0 and aj = 1 , then c = y = 0 . 
xy — 0 and x — 0, then d = 5 . 
= ocy x 0(1 - + 5 (1 _ xy)(l - x) , 
and also xy( 1 — a?) = 0 , as evidently ought to be the case. 
By putting in 
x 2 = x , or x(l - x) = 0 , 
we get v = ^ + 0 a! ( 1 + i7 (1 - *) , 
that is, what is y is identical with what is x and y, together with 
no part of what is x and not y, together with an indefinite part 
of what is not x. The truth of this is evidenced by the diagram. 
Since every logical equation is true arithmetically, 
y - x-y + q (1 r x ) ’ 
where the bar denotes that the arithmetical value of the symbol is 
taken. 
Applications of the above Theorem. 
(1.) The probability that it thunders upon a given day is^, the 
probability that it both thunders and hails is q, but of the connec- 
tion of the two phenomena of thunder and hail, nothing further is 
supposed to be known. Required the probability that it hails on 
the proposed day. 
