diff --git a/Exercises/Apostol/Exercises_1_7.tex b/Exercises/Apostol/Exercises_1_7.tex
index 51268f3..4b0b4a5 100644
--- a/Exercises/Apostol/Exercises_1_7.tex
+++ b/Exercises/Apostol/Exercises_1_7.tex
@@ -1,7 +1,10 @@
 \documentclass{article}
 \usepackage{amsmath}
+\usepackage{graphicx}
+\usepackage{mathrsfs}
 
 \input{../../preamble}
+\graphicspath{{./images/}}
 
 \newcommand{\larea}[2]{\lean{../..}{Bookshelf/Real/Geometry/Area}{#1}{#2}}
 \newcommand{\lrect}[2]{\lean{../..}{Bookshelf/Real/Geometry/Rectangle}{#1}{#2}}
@@ -9,7 +12,7 @@
 \begin{document}
 
 The properties of area in this set of exercises are to be deduced from the
-axioms for area stated in the foregoing section.
+  axioms for area stated in the foregoing section.
 
 \section{Exercise 1}%
 \label{sec:exercise-1}
@@ -40,20 +43,20 @@ A set consisting of a finite number of points in a plane.
 
 \begin{proof}
 
-  We show for all $k > 0$, a set consisting of $k$ points in a plane is
-  measurable with area $0$.
+  Define predicate $P(n)$ as "A set consisting of $n$ points in a plane is
+    measurable with area $0$".
+  We use induction to prove $P(n)$ holds for all $n > 0$.
 
   \paragraph{Base Case}%
 
     Consider a set $S$ consisting of a single point in a plane.
     By \eqref{sub:exercise-1a}, $S$ is measurable with area $0$.
+    Thus $P(1)$ holds.
 
   \paragraph{Induction Step}%
 
-    Define our induction hypothesis as, "Let $k > 0$ and assume a set consisting
-      of $k$ points in a plane is measurable with area $0$."
-
-    Consider a set $S_{k+1}$ consisting of $k + 1$ points in a plane.
+    Assume induction hypothesis $P(k)$ holds for some $k > 0$.
+    Let $S_{k+1}$ be a set consisting of $k + 1$ points in a plane.
     Pick an arbitrary point of $S_{k+1}$.
     Denote the set containing just this point as $T$.
     Denote the remaining set of points as $S_k$.
@@ -68,8 +71,6 @@ A set consisting of a finite number of points in a plane.
           & = a(S_k) + a(T) - a(S_k \cap T) \nonumber \\
           & = 0 + 0 - a(S_k \cap T). \label{sub:exercise-1b-eq1}
       \end{align}
-
-    \noindent
     There are two cases to consider:
 
     \subparagraph{Case 1}%
@@ -86,12 +87,11 @@ A set consisting of a finite number of points in a plane.
     \vspace{8pt}
     \noindent
     In both cases, \eqref{sub:exercise-1b-eq1} evaluates to $0$, implying
-      $a(S_{k+1}) = 0$ as expected.
+      $P(k + 1)$ as expected.
 
   \paragraph{Conclusion}%
 
-    By mathematical induction, it follows for all $n > 0$, a set consisting of
-    $n$ points in a plane is measurable with area $0$.
+    By mathematical induction, it follows for all $n > 0$, $P(n)$ is true.
 
 \end{proof}
 
@@ -102,8 +102,9 @@ The union of a finite collection of line segments in a plane.
 
 \begin{proof}
 
-  We show for all $k > 0$, a set consisting of $k$ line segments in a plane is
-  measurable with area $0$.
+  Define predicate $P(n)$ as "A set consisting of $n$ line segments in a plane
+    is measurable with area $0$".
+  We use induction to prove $P(n)$ holds for all $n > 0$.
 
   \paragraph{Base Case}%
 
@@ -113,13 +114,12 @@ The union of a finite collection of line segments in a plane.
     By \larea{Choice-of-Scale}{Choice of Scale}, $S$ is measurable with area its
       width $w$ times its height $h$.
     Therefore $a(S) = wh = 0$.
+    Thus $P(1)$ holds.
 
   \paragraph{Induction Step}%
 
-    Define our induction hypothesis as, "Let $k > 0$ and assume a set consisting
-      of $k$ line segments in a plane is measurable with area $0$."
-
-    Consider a set $S_{k+1}$ consisting of $k + 1$ line segments in a plane.
+    Assume induction hypothesis $P(k)$ holds for some $k > 0$.
+    Let $S_{k+1}$ be a set consisting of $k + 1$ line segments in a plane.
     Pick an arbitrary line segment of $S_{k+1}$.
     Denote the set containing just this line segment as $T$.
     Denote the remaining set of line segments as $S_k$.
@@ -134,8 +134,6 @@ The union of a finite collection of line segments in a plane.
           & = a(S_k) + a(T) - a(S_k \cap T) \nonumber \\
           & = 0 + 0 - a(S_k \cap T). \label{sub:exercise-1c-eq1}
       \end{align}
-
-    \noindent
     There are two cases to consider:
 
     \subparagraph{Case 1}%
@@ -152,12 +150,11 @@ The union of a finite collection of line segments in a plane.
     \vspace{8pt}
     \noindent
     In both cases, \eqref{sub:exercise-1c-eq1} evaluates to $0$, implying
-      $a(S_{k+1}) = 0$ as expected.
+      $P(k + 1)$ as expected.
 
   \paragraph{Conclusion}%
 
-    By mathematical induction, it follows for all $n > 0$, a set consisting of
-    $n$ line segments in a plane is measurable with area $0$.
+    By mathematical induction, it follows for all $n > 0$, $P(n)$ is true.
 
 \end{proof}
 
@@ -165,12 +162,450 @@ The union of a finite collection of line segments in a plane.
 \label{sec:exercise-2}
 
 Every right triangular region is measurable because it can be obtained as the
-intersection of two rectangles. Prove that every triangular region is measurable
-and that its area is one half the product of its base and altitude.
+  intersection of two rectangles.
+Prove that every triangular region is measurable and that its area is one half
+  the product of its base and altitude.
 
 \begin{proof}
 
-  TODO
+  Let $T'$ be a triangular region with base of length $a$, height of length $b$,
+    and hypotenuse of length $c$.
+  Consider the translation and rotation of $T'$, say $T$, such that its
+    hypotenuse is entirely within quadrant I and the vertex opposite the
+    hypotenuse is situated at point $(0, 0)$.
+
+  Let $R$ be a rectangle of width $a$, height $b$, and bottom-left corner at
+    $(0, 0)$.
+  By construction, $R$ covers all of $T$.
+  Let $S$ be a rectangle of width $c$ and height $a\sin{\theta}$, where $\theta$
+    is the acute angle measured from the bottom-right corner of $T$ relative
+    to the $x$-axis.
+  As an example, consider the image below of triangle $T$ with width $4$ and
+    height $3$:
+
+  \begin{figure}[h]
+    \includegraphics{right-triangle}
+    \centering
+  \end{figure}
+
+  By \larea{Choice-of-Scale}{Choice of Scale}, both $R$ and $S$ are measurable.
+  By this same axiom, $a(R) = ab$ and $a(S) = ca\sin{\theta}$.
+  By the \larea{Additive-Property}{Additive Property}, $R \cup S$ and $R \cap S$
+    are both measurable.
+  $a(R \cap S) = a(T)$ and $a(R \cup S)$ can be determined by noting that
+    $R$'s construction implies identity $a(R) = 2a(T)$.
+  Therefore
+    \begin{align*}
+      a(T)
+        & = a(R \cap S) \\
+        & = a(R) + a(S) - a(R \cup S) \\
+        & = ab + ca\sin{\theta} - a(R \cup S) \\
+        & = ab + ca\sin{\theta} - (ca\sin{\theta} + \frac{1}{2}a(R)) \\
+        & = ab + ca\sin{\theta} - ca\sin{\theta} - a(T).
+    \end{align*}
+  Solving for $a(T)$ gives the desired identity: $$a(T) = \frac{1}{2}ab.$$
+  By \larea{Invariance-Under-Congruence}{Invariance Under Congruence},
+    $a(T') = a(T)$, concluding our proof.
+
+\end{proof}
+
+\section{Exercise 3}%
+\label{sec:exercise-3}
+
+Prove that every trapezoid and every parallelogram is measurable and derive the
+  usual formulas for their areas.
+
+\begin{proof}
+
+  We begin by proving the formula for a trapezoid.
+  Let $S$ be a trapezoid with height $h$ and bases $b_1$ and $b_2$, $b_1 < b_2$.
+  There are three cases to consider:
+
+  \begin{figure}[h]
+    \includegraphics[width=\textwidth]{trapezoid-cases}
+    \centering
+  \end{figure}
+
+  \paragraph{Case 1}%
+
+    Suppose $S$ is a right trapezoid.
+    Then $S$ is the union of non-overlapping rectangle $R$ of width $b_1$ and
+      height $h$ with right triangle $T$ of base $b_2 - b_1$ and height $h$.
+    By \larea{Choice-of-Scale}{Choice of Scale}, $R$ is measurable.
+    By \eqref{sec:exercise-2}, $T$ is measurable.
+    By the \larea{Additive-Property}{Additive Property}, $R \cup T$ and $R \cap T$
+      are both measurable and
+      \begin{align*}
+        a(S)
+          & = a(R \cup T) \\
+          & = a(R) + a(T) - a(R \cap T) \\
+          & = a(R) + a(T) & \text{by construction} \\
+          & = b_1h + a(T) & \text{Choice of Scale} \\
+          & = b_1h + \frac{1}{2}(b_2 - b_1)h & \eqref{sec:exercise-2} \\
+          & = \frac{b_1 + b_2}{2}h.
+      \end{align*}
+
+  \paragraph{Case 2}%
+
+    Suppose $S$ is an acute trapezoid.
+    Then $S$ is the union of non-overlapping triangle $T$ and right trapezoid $R$.
+    Let $c$ denote the length of base $T$.
+    Then $R$ has longer base edge of length $b_2 - c$.
+    By \eqref{sec:exercise-2}, $T$ is measurable.
+    By Case 1, $R$ is measurable.
+    By the \larea{Additive-Property}{Additive Property}, $R \cup T$ and $R \cap T$
+      are both measurable and
+      \begin{align*}
+        a(S)
+          & = a(T) + a(R) - a(R \cap T) \\
+          & = a(T) + a(R) & \text{by construction} \\
+          & = \frac{1}{2}ch + a(R) & \eqref{sec:exercise-2} \\
+          & = \frac{1}{2}ch + \frac{b_1 + b_2 - c}{2}h & \text{Case 1} \\
+          & = \frac{b_1 + b_2}{2}h.
+      \end{align*}
+
+  \paragraph{Case 3}%
+
+    Suppose $S$ is an obtuse trapezoid.
+    Then $S$ is the union of non-overlapping triangle $T$ and right trapezoid $R$.
+    Let $c$ denote the length of base $T$.
+    Reflect $T$ vertically to form another right triangle, say $T'$.
+    Then $T' \cup R$ is an acute trapezoid.
+    By \larea{Invariance-Under-Congruence}{Invariance Under Congruence},
+      \begin{equation}
+        \label{par:exercise-3-case-3-eq1}
+        \tag{3.1}
+        a(T' \cup R) = a(T \cup R).
+      \end{equation}
+    By construction, $T' \cup R$ has height $h$ and bases $b_1 - c$ and $b_2 + c$
+      meaning
+      \begin{align*}
+        a(T \cup R)
+          & = a(T' \cup R) & \eqref{par:exercise-3-case-3-eq1} \\
+          & = \frac{b_1 - c + b_2 + c}{2}h & \text{Case 2} \\
+          & = \frac{b_1 + b_2}{2}h.
+      \end{align*}
+
+  \paragraph{Conclusion}%
+
+    These cases are exhaustive and in agreement with one another.
+    Thus $S$ is measurable and $$a(S) = \frac{b_1 + b_2}{2}h.$$
+
+  \vspace{4pt}
+  \hrule
+  \vspace{10pt}
+
+  Let $P$ be a parallelogram with base $b$ and height $h$.
+  Then $P$ is the union of non-overlapping triangle $T$ and right trapezoid $R$.
+  Let $c$ denote the length of base $T$.
+  Reflect $T$ vertically to form another right triangle, say $T'$.
+  Then $T' \cup R$ is an acute trapezoid.
+  By \larea{Invariance-Under-Congruence}{Invariance Under Congruence},
+    \begin{equation}
+      \label{par:exercise-3-eq2}
+      \tag{3.2}
+      a(T' \cup R) = a(T \cup R).
+    \end{equation}
+  By construction, $T' \cup R$ has height $h$ and bases $b - c$ and $b + c$
+    meaning
+    \begin{align*}
+      a(T \cup R)
+        & = a(T' \cup R) & \eqref{par:exercise-3-eq2} \\
+        & = \frac{b - c + b + c}{2}h & \text{Area of Trapezoid} \\
+        & = bh.
+    \end{align*}
+
+\end{proof}
+
+\section{Exercise 4}%
+\label{sec:exercise-4}
+
+A point $(x, y)$ is called a \textit{lattice point} if both coordinates $x$ and
+  $y$ are integers.
+Let $P$ be a polygon whose vertices are lattice points.
+The area of $P$ is $I + \frac{1}{2}B - 1$, where $I$ denotes the number of
+  lattice points inside the polygon and $B$ denotes the number on the boundary.
+
+\subsection{Exercise 4a}%
+\label{sub:exercise-4a}
+
+Prove that the formula is valid for rectangles with sides parallel to the
+  coordinate axes.
+
+\begin{proof}
+
+  Let $P$ be a rectangle with width $w$, height $h$, and lattice points for
+    vertices.
+  We assume $P$ has three non-collinear points, ruling out any instances of
+    points or line segments.
+  By \larea{Choice-of-Scale}{Choice of Scale}, $P$ is measurable with area
+    $a(P) = wh$.
+  By construction, $P$ has $I = (w - 1)(h - 1)$ interior lattice points and
+    $B = 2(w + h)$ lattice points on its boundary.
+  The following shows the lattice point area formula is in agreement with
+    $a(P)$:
+    \begin{align*}
+      I + \frac{1}{2}B - 1
+        & = (w - 1)(h - 1) + \frac{1}{2}\left[ 2(w + h) \right] - 1 \\
+        & = (wh - w - h + 1) + \frac{1}{2}\left[ 2(w + h) \right] - 1 \\
+        & = (wh - w - h + 1) + (w + h) - 1 \\
+        & = wh.
+    \end{align*}
+
+\end{proof}
+
+\subsection{Exercise 4b}%
+\label{sub:exercise-4b}
+
+Prove that the formula is valid for right triangles and parallelograms.
+
+\begin{proof}
+
+  Let $T'$ be a right triangle with width $w$ and height $h$.
+  Let $T$ be the triangle $T'$ translated, rotated, and reflected such that the
+    its vertices are $(0, 0)$, $(0, w)$, and $(w, h)$, where $w \leq h$.
+  Let $R$ be the overlapping rectangle of width $w$ and height $h$, situated
+    with bottom-left corner at $(0, 0)$.
+  There are two cases to consider:
+
+  \paragraph{Case 1}%
+
+    Suppose $h / w$ is an integral value.
+    Then there exist $w + 1$ lattice points on $T$'s hypotenuse.
+    The number of interior lattices points of $T$ is
+      \begin{align*}
+        I
+          & = \frac{1}{2}\left[ (w - 1)(h - 1) - (w - 1) \right] \\
+          & = \frac{1}{2}\left[ wh - 2w - h + 2 \right].
+      \end{align*}
+    The number of boundary lattice points of $T$ is
+      \begin{align*}
+        B
+          & = (w + 1) + h + (w - 1) \\
+          & = 2w + h.
+      \end{align*}
+    Thus
+      \begin{align*}
+        I + \frac{1}{2}B - 1
+          & = \frac{wh - 2w - h + 2}{2} + \frac{2w + h}{2} - 1 \\
+          & = \frac{wh - 2w - h + 2 + 2w + h - 2}{2} \\
+          & = \frac{wh}{2}.
+      \end{align*}
+
+  \paragraph{Case 2}%
+
+    Suppose $h / w$ is not an integral value.
+    Then there exist exactly 2 lattice points on $T$'s hypotenuse.
+    The number of interior lattice points of $T$ is
+      \begin{align*}
+        I
+          & = \frac{1}{2}\left[ (w - 1)(h - 1) \right] \\
+          & = \frac{1}{2}\left[ wh - w - h + 1 \right].
+      \end{align*}
+    The number of boundary lattice points of $T$ is
+      \begin{align*}
+        B
+          & = (w + 1) + h \\
+          & = w + h + 1.
+      \end{align*}
+    Thus
+      \begin{align*}
+        I + \frac{1}{2}B - 1
+          & = \frac{wh - w - h + 1}{2} + \frac{w + h + 1}{2} - 1 \\
+          & = \frac{wh - w - h + 1 + w + h + 1 - 2}{2} \\
+          & = \frac{wh}{2}.
+      \end{align*}
+
+  \paragraph{Conclusion}%
+
+    These cases are exhaustive and in agreement with one another.
+    Thus $$a(T) = I + \frac{1}{2}B - 1.$$
+    We do not prove this formula is valid for parallelograms here.
+    Instead, refer to \eqref{sub:exercise-4c} below.
+
+\end{proof}
+
+\subsection{Exercise 4c}%
+\label{sub:exercise-4c}
+
+Use induction on the number of edges to construct a proof for general polygons.
+
+\begin{proof}
+
+  Define predicate $P(n)$ as "An $n$-polygon with vertices on lattice points has
+    area $I + \frac{1}{2}B - 1$."
+  We use induction to prove $P(n)$ holds for all $n \geq 3$.
+
+  \paragraph{Base Case}%
+
+    A $3$-polygon is a triangle.
+    By \eqref{sub:exercise-4b}, the lattice point area formula holds.
+    Thus $P(3)$ holds.
+
+  \paragraph{Induction Step}%
+
+    Assume induction hypothesis $P(k)$ holds for some $k \geq 3$.
+    Let $P$ be a $(k + 1)$-polygon with vertices on lattice points.
+    Such a polygon is equivalent to the union of a $k$-polygon $S$ with a
+      triangle $T$.
+    That is, $P = S \cup T$.
+
+    Let $I_P$ be the number of interior lattice points of $P$.
+    Let $B_P$ be the number of boundary lattice points of $P$.
+    Similarly, let $I_S$, $I_T$, $B_S$, and $B_T$ be the number of interior
+      and boundary lattice points of $S$ and $T$.
+    Let $c$ denote the number of boundary points shared between $S$ and $T$.
+
+    By our induction hypothesis, $a(S) = I_S + \frac{1}{2}B_S - 1$.
+    By our base case, $a(T) = I_T + \frac{1}{2}B_T - 1$.
+    By construction, it follows:
+      \begin{align*}
+        I_P & = I_S + I_T + c - 2 \\
+        B_P & = B_S + B_T - (c - 2) - c \\
+            & = B_S + B_T - 2c + 2.
+      \end{align*}
+    Applying the lattice point area formula to $P$ yields the following:
+      \begin{align*}
+        & I_P + \frac{1}{2}B_P - 1 \\
+          & = (I_S + I_T + c - 2) + \frac{1}{2}(B_S + B_T - 2c + 2) - 1 \\
+          & = I_S + I_T + c - 2 + \frac{1}{2}B_S + \frac{1}{2}B_T - c + 1 - 1 \\
+          & = (I_S + \frac{1}{2}B_S - 1) + (I_T + \frac{1}{2}B_T - 1) \\
+          & = a(S) + (I_T + \frac{1}{2}B_T - 1) & \text{induction hypothesis} \\
+          & = a(S) + a(T). & \text{base case}
+      \end{align*}
+    By the \larea{Additive-Property}{Additive Property}, $S \cup T$ is
+      measurable, $S \cap T$ is measurable, and
+      \begin{align*}
+        a(P)
+          & = a(S \cup T) \\
+          & = a(S) + a(T) - a(S \cap T) \\
+          & = a(S) + a(T). & \text{by construction}
+      \end{align*}
+    This shows the lattice point area formula is in agreement with our axiomatic
+      definition of area.
+    Thus $P(k + 1)$ holds.
+
+  \paragraph{Conclusion}%
+
+    By mathematical induction, it follows for all $n \geq 3$, $P(n)$ is true.
+
+\end{proof}
+
+\subsection{Exercise 5}%
+\label{sub:exercise-5}
+
+Prove that a triangle whose vertices are lattice points cannot be equilateral.
+
+[\textit{Hint:} Assume there is such a triangle and compute its area in two
+ways, using Exercises 2 and 4.]
+
+\begin{proof}
+
+  Proceed by contradiction.
+  Let $T$ be an equilateral triangle whose vertices are lattice points.
+  Assume each side of $T$ has length $a$.
+  Then $T$ has height $h = (a\sqrt{3}) / 2$.
+  By \eqref{sec:exercise-2},
+    \begin{equation}
+      \label{sub:exercise-5-eq1}
+      \tag{5.1}
+      a(T) = \frac{1}{2}ah = \frac{a^2\sqrt{3}}{4}.
+    \end{equation}
+  Let $I$ and $B$ denote the number of interior and boundary lattice points of
+    $T$ respectively.
+  By \eqref{sec:exercise-4},
+    \begin{equation}
+      \label{sub:exercise-5-eq2}
+      \tag{5.2}
+      a(T) = I + \frac{1}{2}B - 1.
+    \end{equation}
+  But \eqref{sub:exercise-5-eq1} is irrational whereas
+    \eqref{sub:exercise-5-eq2} is not.
+  This is a contradiction.
+  Thus, there is \textit{no} equilateral triangle whose vertices are lattice
+    points.
+
+\end{proof}
+
+\subsection{Exercise 6}%
+\label{sub:exercise-6}
+
+Let $A = \{1, 2, 3, 4, 5\}$, and let $\mathscr{M}$ denote the class of all
+  subsets of $A$.
+(There are 32 altogether, counting $A$ itself and the empty set $\emptyset$.)
+For each set $S$ in $\mathscr{M}$, let $n(S)$ denote the number of distinct
+  elements in $S$.
+If $S = \{1, 2, 3, 4\}$ and $T = \{3, 4, 5\}$, compute $n(S \cup T)$,
+  $n(S \cap T)$, $n(S - T)$, and $n(T - S)$.
+Prove that the set function $n$ satisfies the first three axioms for area.
+
+\begin{proof}
+
+  Let $S = \{1, 2, 3, 4\}$ and $T = \{3, 4, 5\}$.
+  Then
+    \begin{align*}
+      n(S \cup T)
+        & = n(\{1, 2, 3, 4\} \cup \{3, 4, 5\}) \\
+        & = n(\{1, 2, 3, 4, 5\}) \\
+        & = 5. \\
+      n(S \cap T)
+        & = n(\{1, 2, 3, 4\} \cap \{3, 4, 5\}) \\
+        & = n(\{3, 4\}) \\
+        & = 2. \\
+      n(S - T)
+        & = n(\{1, 2, 3, 4\} - \{3, 4, 5\}) \\
+        & = n(\{1, 2\}) \\
+        & = 2. \\
+      n(T - S)
+        & = n(\{3, 4, 5\} - \{1, 2, 3, 4\}) \\
+        & = n(\{5\}) \\
+        & = 1.
+    \end{align*}
+  We now prove $n$ satisfies the first three axioms for area.
+
+  \paragraph{Nonnegative Property}%
+
+    $n$ returns the length of some member of $\mathscr{M}$.
+    By hypothesis, the smallest possible input to $n$ is $\emptyset$.
+    Since $n(\emptyset) = 0$, it follows $n(S) \geq 0$ for all $S \subset A$.
+
+  \paragraph{Additive Property}%
+
+    Let $S$ and $T$ be members of $\mathscr{M}$.
+    It trivially follows that both $S \cup T$ and $S \cap T$ are in
+      $\mathscr{M}$.
+    Consider the value of $n(S \cup T)$.
+    There are two cases to consider:
+
+    \subparagraph{Case 1}%
+
+      Suppose $S \cap T = \emptyset$.
+      That is, there is no common element shared between $S$ and $T$.
+      Thus
+        \begin{align*}
+          n(S \cup T)
+            & = n(S) + n(T) \\
+            & = n(S) + n(T) - 0 \\
+            & = n(S) + n(T) - n(S \cap T).
+        \end{align*}
+
+    \subparagraph{Case 2}%
+
+      Suppose $S \cap T \neq \emptyset$.
+      Then $n(S) + n(T)$ counts each element of $S \cap T$ twice.
+      Therefore $n(S \cup T) = n(S) + n(T) - n(S \cap T)$.
+
+    \subparagraph{Conclusion}%
+
+      These cases are exhaustive and in agreement with one another.
+      Thus $n(S \cup T) = n(S) + n(T) - n(S \cap T)$.
+
+  \paragraph{Difference Property}%
+
+    Suppose $S, T \in \mathscr{M}$ such that $S \subseteq T$.
+    That is, every member of $S$ is a member of $T$.
+    By definition, $T - S$ consists of members in $T$ but not in $S$.
+    Thus $n(T - S) = n(T) - n(S)$.
 
 \end{proof}
 
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