diff --git a/component_failure_modes_definition/component_failure_modes_definition.tex b/component_failure_modes_definition/component_failure_modes_definition.tex
index 3cabc6a..61147a6 100644
--- a/component_failure_modes_definition/component_failure_modes_definition.tex
+++ b/component_failure_modes_definition/component_failure_modes_definition.tex
@@ -309,10 +309,11 @@ It is an implied requirement of EN298 for instance to consider double simultaneo
 To generalise, we may need to consider $N$ simultaneous 
 failure modes when analysing a functional group. This involves finding
 all combinations of failures modes of size $N$ and less.
-The Powerset concept from Set theory when applied to a set S is the set of all subsets of S, including the empty set and S itself.
+The Powerset concept from Set theory when applied to a set S is the set of all subsets of S, including the empty set and S itself
+\footnote{The empty set is a special case for FMMD analysis, it simply means there
+is no fault active in the functional~group under analysis}.
 In order to consider combinations for the set S where the number of elements in each sub-set of S is $N$ or less, a concept of the `cardinality constrained powerset'
-is proposed and described in the next section. The empty set is a special case for FMMD analysis, it simply means there
-is no fault active in the functional~group under analysis.
+is proposed and described in the next section. 
  
 \subsection{Cardinality Constrained Powerset }
 \label{ccp}
@@ -320,11 +321,16 @@ is no fault active in the functional~group under analysis.
 A Cardinality Constrained powerset is one where sub-sets of a cardinality greater than a threshold
 are not included. This theshold is called the cardinality constraint.
 To indicate this the cardinality constraint $cc$, is subscripted to the powerset symbol thus $\mathcal{P}_{cc}$.
-Consider the set $S = \{a,b,c\}$. $\mathcal{P}_{2} S $ means all subsets of S where the cardinality of the subsets is 
-less than or equal to 2 or less.
+Consider the set $S = \{a,b,c\}$. 
+
+The powerset of S: 
 
 $$ \mathcal{P} S = \{ 0, \{a,b,c\}, \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} $$
 
+
+$\mathcal{P}_{2} S $ means all subsets of S where the cardinality of the subsets is 
+less than or equal to 2 or less.
+
 $$ \mathcal{P}_{2} S = \{ \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} $$
 
 Note that $\mathcal{P}_{1} S $ for this example is:
@@ -352,11 +358,12 @@ from $1$ to $cc$ thus
 
 \begin{equation}
  \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!}    
+ \label{eqn:ccps}
 \end{equation}
 
 
 
-\subsection{Actual Number of combinations to check with Unitary State Fault mode sets}
+\subsection{Actual Number of combinations to check \\ with Unitary State Fault mode sets}
 
 Where all components analysed only have one fault mode, the cardinality constrained powerset
 calculation give the correct number of test case combinations to check.
@@ -367,29 +374,53 @@ be less.
 What must actually be done is to subtract the number of component `internal combinations'
 from the cardinality constrain powerset number.
 
-Thus were we to have a simple circuit with two components R and T, of which
-$FM(R) = {R_o, R_s}$ and $FM(T) = {T_o, T_s, T_h}$.
+Thus were we to have a simple functional group with two components R and T, of which
+$$FM(R) = \{R_o, R_s\}$$ and $$FM(T) = \{T_o, T_s, T_h\}$$.
 For a cardinality constrained powerset of 2, because there are 5 error modes
-gives 
+applying equation \ref{eqn:ccps} gives :-
 
 $$\frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!}  = 15$$
 
-This is composed of
-5 single fault modes, and ${2 \choose 5}$ ten double fault modes.
+This is composed of ${1 \choose 5}$
+five single fault modes, and ${2 \choose 5}$ ten double fault modes.
 However we know that the faults are mutually exclusive for a component.
-We must then subtract the number of `internal' component fault combinations.
+We must then subtract the number of `internal' component fault combinations for each component in the functional~group.
 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 $(3+1)$.
+The number of combinations to check is thus 11 for this example and this can be verified
+by listing all the required combinations:
 
-Written as a general formula, where C is a set of the components (indexed by j where J 
-is the set of componets  in the functional~group under analyis) and $\#C$
+\vbox{
+%\tiny
+\begin{enumerate}
+\item $\{R_o T_o\}$
+\item $\{R_o T_s\}$
+\item $\{R_o T_h\}$
+\item $\{R_s T_o\}$
+\item $\{R_s T_s\}$
+\item $\{R_s T_h\}$
+\item $\{R_o \}$
+\item $\{R_s \}$
+\item $\{T_o \}$
+\item $\{T_s \}$
+\item $\{T_h \}$
+\end{enumerate}
+%\normalsize
+}
+
+
+The cardinality constrained powerset equation \ref{eqn:ccps} corrected for 
+unitary state failure modes can be 
+written as a general formula, where C is a set of the components (indexed by j where J 
+is the set of components  in the functional~group under analyis) and $\#C$
 indicates the number of mutually exclusive fault modes each component has:-
 
 %$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!}    $$
 \begin{equation}
  \#\mathcal{P}_{cc} S = {\sum^{k}_{1..cc}  \frac{\#S!}{k!(\#S-k)!}}  - {\sum^{j}_{j \in J} {\#C_{j} \choose cc}} 
+ \label{eqn:correctedccps}
 \end{equation}
 
 
@@ -416,7 +447,7 @@ $$ F = \Omega(C) \backslash OK $$
 
 The $OK$ statistical case is the largest in probability, and is therefore
 of interest when analysing systems from a statistical perspective.
-This is of interest to conditional probability calculations
+This is of interest for the application of conditional probability calculations
 such as Bayes theorem.
 
 \vspace{40pt}