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Add If and Only If Proofs (#7)
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3 changes: 3 additions & 0 deletions .vscode/ltex.hiddenFalsePositives.en-GB.txt
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{"rule":"MORFOLOGIK_RULE_EN_GB","sentence":"^\\QPackages Used\nTeX packages that have been used are:\namsmath, amssymb: Used for mathematical symbols and environments\namsthm: Used for the proof environment\\\nbabel: Use description environment, and chapter renaming\nchangepage: Allow paragraphs to be indented\nenumitem: Used to improve the way lists are displayed\netoolbox: Allow commands to be patched\nfancyhdr: Improve header and footer display\nfontenc, charter, inconsolata, cabin, newtxmath, bm: Font display\ngeometry: Used for document margins\nhyperref: Add PDF bookmarks\nparskip: Set proper default values for paragraph separation\nsubfiles: Allow \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q files to be written separately, but built together\ntcolorbox, adjustbox: Create colored boxes for different environments\ntikz: Used for images\nvenndiagram: Easier support for Venn Diagram display than using pure tikz\nxargs: Allow commands and environments to have more than one optional argument\nxcolor: Used to get more colors\nxstring: Analyze strings given as arguments to an environment\\E$"}
{"rule":"MORFOLOGIK_RULE_EN_GB","sentence":"^\\QCommands Added\nFor intellisense, a list of the commands is added in the package_intellisense folder.\\E$"}
{"rule":"POSSESSIVE_APOSTROPHE","sentence":"^\\QIf you want to access pre-generated PDFs, please check the Releases section on the sidebar.\\E$"}
{"rule":"LC_AFTER_PERIOD","sentence":"^\\Q\\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q [above] at (current bounding box.south) \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q; How many learners play Mastermind or basketball, (or both)?\\E$"}
{"rule":"CD_NN","sentence":"^\\Q\\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q A questionnaire filled in by 100 subscribers to Blue Scalpel Medical Insurance who submitted no claims during 2009 reveals that 45 jog regularly, 30 do aerobics regularly, 20 cycle regularly, 6 jog and do aerobics, 1 jogs and cycles, 5 do aerobics and cycle, and 1 jogs, cycles and does aerobics.\\E$"}
{"rule":"LC_AFTER_PERIOD","sentence":"^\\Q\\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q [above] at (current bounding box.south) \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q; How many of these healthy people do not participate regularly in any of the three activities?\\E$"}
141 changes: 141 additions & 0 deletions units/unit04.tex
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\end{minipage}
\end{example}
\pagebreak
\subsection{If and Only If Proofs}
The purpose of an iff proof is to shorten a proof where you need to show that it works both forwards and backwards. Remember the symbol for iff is $\leftrightarrow$.

To do this, you convert the statement into words.

\begin{example}
\begin{questions}[before=\raggedright]
\item Prove that $A \cup (B \cup C) = (A \cup B) \cup C$ for all sets $A$, $B$, $C \subseteq U$.\\
\begin{answer}
\begin{minipage}{0.6\textwidth}
To start off, assume that $x$ is an element of the statement on the left:
\begin{indentparagraph}
Let $x \in A \cup (B \cup C)$.
\end{indentparagraph}
Then start the proof. Convert all the $\cup$ and $\cap$ symbols to words.
\begin{indentparagraph}
\begin{tabbing}
$\qquad$ \= $x \in A \cup (B \cup C)$\\
iff \> $x \in A$ or $x \in (B \cup C)$\\
iff \> $x \in A$ or $x \in B$ or $x \in C$\\
iff \> ($x \in A$ or $x \in B$) or $x \in C$\\
iff \> ($x \in (A \cup B)$) or $x \in C$\\
iff \> $x \in (A \cup B) \cup C$\\
$\therefore$ \> $A \cup (B \cup C) = (A \cup B) \cup C$
\end{tabbing}
\end{indentparagraph}
\end{minipage}
\begin{minipage}{0.33\textwidth}
\begin{center}
\begin{vennthree}[][]
\fillA
\fillB
\fillC
\end{vennthree}
\end{center}
\end{minipage}
\end{answer}
\item Prove that $A \cap (B \cap C) = (A \cap B) \cap C$ for all sets $A$, $B$, $C \subseteq U$.\\
\begin{answer}
\begin{minipage}{0.6\textwidth}
Let $x \in A \cap (B \cap C)$
\begin{tabbing}
$\qquad$ \= $x \in A \cap (B \cap C)$\\
iff \> $x \in A$ and $x \in (B \cap C)$\\
iff \> $x \in A$ and $x \in B$ and $x \in C$\\
iff \> ($x \in A$ and $x \in B$) and $x \in C$\\
iff \> ($x \in (A \cap B)$) and $x \in C$\\
iff \> $x \in (A \cap B) \cap C$\\
$\therefore$ \> $A \cap (B \cap C) = (A \cap B) \cap C$
\end{tabbing}
\end{minipage}
\begin{minipage}{0.33\textwidth}
\begin{center}
\begin{vennthree}[][]
\fillACapBCapC
\end{vennthree}
\end{center}
\end{minipage}
\end{answer}
\end{questions}
\end{example}
\pagebreak
\subsubsection{Using Nots}
A basic application of the not symbol swaps $\in$ to $\notin$. If the statement has a $-$ in, this is the equivalent of and $\notin$. For example,
\begin{align*}
x \in A - B &= x \in A \text{ and } x \notin B
\end{align*}
\begin{example}
\begin{questions}[before=\raggedright]
\item Prove that $(A')' = A$ for all sets $A \subseteq U$.
\begin{answer}
\begin{minipage}{0.6\textwidth}
Let $x \in (A')'$
\begin{tabbing}
$\qquad$ \= $x \in (A')'$\\
iff \> $x \notin A'$\\
iff \> $x$ is not $\notin A$\\
iff \> $x \in A$\\
$\therefore$ \> $(A')' = A$
\end{tabbing}
\end{minipage}
\begin{minipage}{0.33\textwidth}
\begin{center}
\begin{venntwo}[][]
\fillA
\end{venntwo}
\end{center}
\end{minipage}
\end{answer}
\end{questions}
\end{example}
\begin{sidenote}{The words swap!}
When you apply a $\notin$ sign in words, then $\cup$ means \emph{and} instead of or, and $\cap$ means \emph{or} instead of and.
\end{sidenote}
\begin{example}
\begin{questions}[before=\raggedright]
\item Prove that $(A \cup B)' = A' \cap B'$.\\
\begin{answer}
\begin{minipage}{0.6\textwidth}
Let $x \in (A \cup B)'$.
\begin{tabbing}
$\qquad$ \= $x \in (A \cup B)'$\\
iff \> $x \notin (A \cup B)$\\
iff \> $x \notin A$ \emph{and} $x \notin B$ $\quad$ (and instead of or for $\cup$)\\
iff \> $x \in A'$ and $x \in B'$\\
iff \> $x \in A' \cap B'$\\
$\therefore$ \> $(A \cup B)' = A' \cap B'$
\end{tabbing}
\end{minipage}
\begin{minipage}{0.33\textwidth}
\begin{center}
\begin{venntwo}[][]
\fillNotAorB
\end{venntwo}
\end{center}
\end{minipage}
\end{answer}
\item Prove that $(A \cap B)' = A' \cup B'$.\\
\begin{answer}
\begin{minipage}{0.6\textwidth}
Let $x \in (A \cap B)'$.
\begin{tabbing}
$\qquad$ \= $x \in (A \cap B)'$\\
iff \> $x \notin (A \cap B)$\\
iff \> $x \notin A$ \emph{or} $x \notin B$ $\quad$ (or instead of and for $\cap$)\\
iff \> $x \in A'$ or $x \in B'$\\
iff \> $x \in A' \cup B'$\\
$\therefore$ \> $(A \cap B)' = A' \cup B'$
\end{tabbing}
\end{minipage}
\begin{minipage}{0.33\textwidth}
\begin{center}
\begin{venntwo}[][]
\fillNotA
\fillNotB
\end{venntwo}
\end{center}
\end{minipage}
\end{answer}
\end{questions}
\end{example}
\begin{exercise}{Self-Assessment Exercise \thechapter.6}
\begin{multicols}{2}
\begin{questions}[label=(\alph*), itemsep=0.5em]
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