1185 
first term is positive, then we can show that the two reaction- 
equations, taken in the given or in opposite order, satisfy condition (3). 
Considering the signs of the phases in (41) and in (50°), then it 
is apparent that the phases form in (50°) the same groups as in 
(41); only the group /...#,..#: makes an exception; this is viz. 
separated into two groups, of which the one viz. /’,../-is found at 
the beginning and the other viz. /... HA at the end of (50°). As 
both those groups have, however, an opposite sign, we can again 
unite them to one group. Consequently we find in (41) and in (50°) 
the same groups and with respect to one another in such an order 
of succession, that the series of signs of (41) and that of (50°) are 
the same. 
We could put the question why in all considerations the series 
of signs of the reaction-equation : 
a,b, +... + A1 Fr + ap F, +...+ ante Fata ="). (o0e) 
is used and not that of the equation: 
ma, + .. + Up—1dp—1 Fp—1 HW apFp +. + UntogntePnte= 0 (50S) 
We might just as well have used this, for both the reaction- 
equations have the same series of signs. In order to find the series 
of signs of (504), we must give another order of succession to the 
phases, viz. in such a way that condition (3) is satisfied. Now the 
ratios are, however, no more: 
but eel 1 
fis. Were ely 129 WG = ge gts F 
t1 3 {n+ al ls Uno 
When we take uw,..."p—1 positive and w,...Uj42 negative, then 
we find: 
So. > es 
Up t, Un+2 Up 
Hence it is apparent, ae that we have to write the reaction- 
equations : 
Bp—1%)—1F pi al uaf, = Urn Fn? am epo  UpapEp = — 0 (509) 
ap -1F pr ee a,F, 4 NEVE +...4+ ars ==) ae (50%) 
When a, is negative, then we give the opposite sign to all 
phases. Considering the signs of the phases in (50°) and (507) then it 
appears that both equations contain the same groups, so that both 
have the same series of signs. 
It is apparent from our considerations that with each coneentration- 
diagramtype corresponds a P,7-diagramtype and reversally and that 
