diff --git a/component_failure_modes_definition/component_failure_modes_definition.tex b/component_failure_modes_definition/component_failure_modes_definition.tex
index 189b9e4..3cabc6a 100644
--- a/component_failure_modes_definition/component_failure_modes_definition.tex
+++ b/component_failure_modes_definition/component_failure_modes_definition.tex
@@ -302,6 +302,18 @@ we have banned larger combinations as well.
 
 \section{Handling Simultaneous Component Faults}
 
+For some integrity levels of static analysis there is a need to consider not only single
+failure modes in isolation, but cases where more then one failure mode may occur
+simultaneously.
+It is an implied requirement of EN298 for instance to consider double simultaneous faults.
+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.
+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.
+ 
 \subsection{Cardinality Constrained Powerset }
 \label{ccp}
 
@@ -309,14 +321,18 @@ A Cardinality Constrained powerset is one where sub-sets of a cardinality greate
 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.
+less than or equal to 2 or less.
 
 $$ \mathcal{P} S = \{ 0, \{a,b,c\}, \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} $$
 
 $$ \mathcal{P}_{2} S = \{ \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} $$
 
+Note that $\mathcal{P}_{1} S $ for this example is:
+
 $$ \mathcal{P}_{1} S = \{ \{a\},\{b\},\{c\} \} $$
 
+\paragraph{Calculating the number of elements in a cardinality constrained powerset}
+
 A $k$ combination is a subset with $k$ elements.  
 The number of $k$ combinations (each of size $k$) from a set $S$ 
 with $n$ elements (size $n$) is the binomial coefficient
@@ -334,7 +350,9 @@ from $1$ to $cc$ thus
 % $$ {\sum}_{k = 1..cc} {\#S \choose k} = \frac{\#S!}{k!(\#S-k)!} $$
 %
 
-$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!}    $$
+\begin{equation}
+ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!}    
+\end{equation}
 
 
 
@@ -352,7 +370,11 @@ 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 $FM(T) = {T_o, T_s, T_h}$.
 For a cardinality constrained powerset of 2, because there are 5 error modes
-gives $\frac{5!}/{1!(5-1)!} + \frac{5!}{2!(5-2)!}  = 15$. OK
+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.
 However we know that the faults are mutually exclusive for a component.
 We must then subtract the number of `internal' component fault combinations.
@@ -362,12 +384,13 @@ $R_o \wedge R_s$. As a combination ${2 \choose 2} = 1$ . For $T$ the component w
 Thus for $cc == 2$ we must subtract $(3+1)$.
 
 Written as a general formula, where C is a set of the components (indexed by j where J 
-is the set of componets under analyis) and $\#C$
-indicates the number of mutually exclusive fault modes the compoent has:-
+is the set of componets  in the functional~group under analyis) and $\#C$
+indicates the number of mutually exclusive fault modes each component has:-
 
 %$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!}    $$
-
-$$ \#\mathcal{P}_{cc} S = {\sum^{k}_{1..cc}  \frac{\#S!}{k!(\#S-k)!}}  - {\sum^{j}_{j \in J} {\#C_{j} \choose cc}}  $$
+\begin{equation}
+ \#\mathcal{P}_{cc} S = {\sum^{k}_{1..cc}  \frac{\#S!}{k!(\#S-k)!}}  - {\sum^{j}_{j \in J} {\#C_{j} \choose cc}} 
+\end{equation}
 
 
 
@@ -393,6 +416,7 @@ $$ F = \Omega(C) \backslash OK $$
 
 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
+such as Bayes theorem.
 
 \vspace{40pt}