making it clearer

component_failure_modes_definition/component_failure_modes_definition.tex

@@ -377,7 +377,7 @@ from the cardinality constrain powerset number.
 
 \subsubsection{Example: Two Component functional group \\ cardinality Constraint of 2}
 
-Thus were we to have a simple functional group with two components R and T, of which
+For example: were we to have a simple functional group with two components R and T, of which
 $$FM(R) = \{R_o, R_s\}$$ and $$FM(T) = \{T_o, T_s, T_h\}$$.
 
 This means that a functional~group $FG=\{R,T\}$ will have a component failure modes set % $FM_{cfg} $
@@ -395,7 +395,7 @@ We must then subtract the number of `internal' component fault combinations for
 For component R there is only one internal component fault that cannot exist
 $R_o \wedge R_s$. As a combination ${2 \choose 2} = 1$ . For $T$ the component with 
   three  fault modes ${2 \choose 3} = 3$.
-Thus for $cc == 2$ we must subtract $(3+1)$.
+Thus for $cc == 2$, under the conditions of unitary state failure modes in the components $R$ and $T$, we must subtract $(3+1)$.
 The number of combinations to check is thus 11 for this example and this can be verified
 by listing all the required combinations:
 %