518 
In the reaction: 
G+Z,2L+Z,orG+4ZA L—Z=0 
G and L have opposite signs, consequently they must have it also 
in all other reactions. 
When we imagine in fig. 1 (VIII) Z, in the point of intersection 
of the lines GZ, and Z,Z,, then 7, and Zare the indifferent phases. 
In the reaction: 
G+Z,2L+2Z or Gt+Z—L- 4=0 
Z, and L have the same sign; consequently they must have it also 
in all other reactions. 
A property of the singular equilibrium M/ is connected with the 
circumstance of the signs of the two indifferent phases being equal 
or not. We shall divide viz. those equilibria into transformable and 
not-transformable equilibria. We call it viz. transformable when the 
singular equilibrium J/ and the invariant equilibrium may be con- 
verted into one another, we call it not-transformable when this 
conversion cannot occur. 
Let us imagine e.g. in fig. J (VIII) Z, in the point of intersection 
of the lines GZ, and Z,Z,; the indifferent phases are then G and 
L and the singular equilibrium is: 
ME. 
This equilibrium M is then not transformable, viz. it cannot be 
converted into the invariant equilibrium and reversally the invariant 
equilibrium cannot be converted into J/. 
When we imagine in fig. 1 (VIII) Z in the point of intersection 
of the lines GZ, and 7,Z,, then Z, and Z are the indifferent phases 
and the singular equilibrium 
ME 
is transformable. A complex of this singular equilibrium viz. is 
represented by a point of the line GZ,; as this point is situated 
within the quadrangle GZ, Z, L, this complex may be converted 
into the invariant equilibrium. 
A complex of the invariant equilibrium is situated within the 
quadrangle GZ Z, L, when we give to this complex such a com- 
position that it is represented by a point of the line GZ,, then the 
invariant equilibrium may be converted into the singular equilibrium 
M. The equilibrium M is, therefore, transformable. 
Now we shall show: 
When the two indifferent phases have the same sign, then the 
singular equilibrium M is transformable, 
