diff --git a/component_failure_modes_definition/component_failure_modes_definition.tex b/component_failure_modes_definition/component_failure_modes_definition.tex
index cbb2c12..0820baf 100644
--- a/component_failure_modes_definition/component_failure_modes_definition.tex
+++ b/component_failure_modes_definition/component_failure_modes_definition.tex
@@ -249,9 +249,9 @@ take a given component $C$ and return its set of failure modes $F$.
 $$  FM : C \mapsto F $$
 
 \begin{definition}
-We can define a set $U$ which is a set of sets of failure modes, where
+We can define a set $\mathcal{U}$ which is a set of sets of failure modes, where
 the component failure modes in each of its members are unitary~state.
-Thus if the failure modes of $F$ are unitary~state, we can say $F \in U$.
+Thus if the failure modes of $F$ are unitary~state, we can say $F \in \mathcal{U}$.
 \end{definition}
 
 \section{Component failure modes:\\ Unitary State example}
@@ -271,7 +271,7 @@ Thus because both fault modes cannot be active at the same time, the intersectio
 
 $$ R_{SHORTED} \cap R_{OPEN} = \emptyset $$
 therefore
-$$ FM(R) \in U $$
+$$ FM(R) \in \mathcal{U} $$
 
 
 We can make this a general case by taking a set $F$ (where $f_1, f_2 \in F$) representing a collection
@@ -283,7 +283,7 @@ We can say that if any pair of fault modes is active at the same time, then the
 unitary state:
 we state this formally
  \begin{equation}
-  \forall f_1,f_2 \in F \dot ( f_1 \neq f_2 \wedge \mathcal{ACTIVE}({f_1}) \wedge \mathcal{ACTIVE}({f_2}) ) \implies F \not\in U 
+  \forall f_1,f_2 \in F \dot ( f_1 \neq f_2 \wedge \mathcal{ACTIVE}({f_1}) \wedge \mathcal{ACTIVE}({f_2}) ) \implies F \not\in \mathcal{U} 
  \end{equation}
 
 
@@ -293,7 +293,7 @@ we state this formally
 % \end{equation}
 
 That is to say that it is impossible that any pair of failure modes can be active at the same time
-for the failure mode set $F$ to exist in the family of sets $U$.
+for the failure mode set $F$ to exist in the family of sets $\mathcal{U}$.
 Note where that are more than two failure~modes, 
 by banning any pairs from being active at the same time
 we have banned larger combinations as well.
@@ -382,8 +382,8 @@ for each component in the functional group under analysis.
 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} $
-of $FM_{cfg} = \{R_o, R_s, T_o, T_s, T_h\}$
+This means that a functional~group $FG=\{R,T\}$ will have a component failure modes set 
+of $FG_{cfg} = \{R_o, R_s, T_o, T_s, T_h\}$
 
 For a cardinality constrained powerset of 2, because there are 5 error modes ( $|{FG_{cfg}}|=5$),
 applying equation \ref{eqn:ccps} gives :-
@@ -432,9 +432,12 @@ where :
 that are members of the functional group $FG$
 i.e. $ \forall j \in J | C_j \in FG $
 \item Let $|{C}_{j}|$
-indicate the number of mutually exclusive fault modes each component has
-\item Let $SU$ be a set of unitary state failure modes from the functional group
-nder analysis $SU = FM(FG)$
+indicate the number of mutually exclusive fault modes of each component
+\item Let $FG_{cfg}$ be the collection of all failure modes
+from all the components in the functional group. 
+\item Let $SU$ be a set of failure modes from the functional group,
+where all contributing components $C_j$ 
+are guaranteed to be `unitary state' i.e. $(SU = FG_{cfg}) \wedge (\forall j \in J | C_j \in \mathcal{U}) $
 \end{itemize}
 %}