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21
Advanced/5D Chess/main.tex Executable file
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\documentclass[
solutions,
shortwarning
]{../../resources/ormc_handout}
\usepackage{../../resources/macros}
\uptitlel{Advanced 2}
\uptitler{\smallurl{}}
\title{5D Chess}
\subtitle{
Prepared by Mark on \today{} \\
Based on \href{https://steamcommunity.com/sharedfiles/filedetails/?id=2177978276}{\texttt{[this]}}
Steam post.
}
\begin{document}
\maketitle
\input{parts/01 rook}
\end{document}

64
Advanced/5D Chess/parts/01 rook.tex Normal file
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\section{Rook Tactics}
\problem{Rook Tactics I}
At first it seems like there's no useful moves you can make ---
if you check the King in the present with your Rook, it just
walks out of the way. Moving your King into the present or past
seems to do nothing exciting (you've only got one turn and it
never puts the king in check, so it's a non-starter).
What if we think five dimensionally?
We can try moving the rook into the past - but the king on the new timeline just moves out of check.
What we need to realize is - the Rook can't just MOVE into the past. It can also CAPTURE into the past. Can we use this to our advantage? Can we do a kind of check that is impossible in 2D chess?
Just move your rook to the northeast corner.
It checks the king in the past.
(The AI can't branch to get away from it since it
would advance the checked timeline and allow me to
make my checkmating move, and it has no other timelines
it can branch to recede the Past while choosing to not
play in the checking timeline.)
\problem{Rook Tactics II}
This would be checkmate in typical chess.
But thanks to time travel, our opponent has a trick up their sleeve.
Do the standard (as in Chess) mate. Enemy king will branch to remove
their present king from check (they can do it since it's not threatened
in the past in that timeline, so it removes the check condition).
Then do the same move again in the new branch. Too bad they couldn't run further!
\problem{Rook Tactics III}
In the last puzzle we saw travel into the past.
But there is one more dimension that pieces can move and attack along.
Move a rook fully north to check a king across timelines.
(The AI can't do anything here except move its pawn -
it again has no other timelines available to branch to recede The Present.
We weren't obliged to play on the future timeline, and in fact if we do
the AI can get out of it by sending their king into the past to capture
our rook. So we have to make a move and NOT make another move.)
\problem{Rook Tactics IV}
Rook Tactics IV: Some things we can try that don't work: The black rooks can make moves into the past and across timelines, but don't result in check.
Also doesn't work: We can move the top board black rook to check a present king, but the present king can just move, so this this leads to nothing good.
So instead, what if we move the top board black rook onto the position a white king used to be in? Then it threatens check by moving into the past to capture the past king.
But the white king in the present can capture the black rook.
Therefore the solution is... To move the bottom board black rook into the same rank and file as the bottom board black rook! Then the top board black rook is protected and can't be captured by a king (it would be captured back by the bottom board black rook). This is checkmate.
\problem{Rook Taactics V}
Do the standard Chess mate. Again, enemy king will branch to remove their present king from check.
Now what? We can check the king with the rook in the new branch, but surely it'll just get captured by the king to stop mate, right? Are we meant to do something else instead?
(One last hint: Pieces can be defended from other timelines.)
Nope! Just move the rook to the same position. Well, why does it work? Because they can't branch to remove mate (two kings AND it'd be an inactive timeline and no other timelines to branch instead), and if they capture the rook instead, your rook from a parallel timeline can capture back. Cross-timeline defense! (If you're unconvinced, try making the capture as black and the game will point it out to you.)

75
Advanced/Definable Sets/parts/0 logic.tex
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@ -1,7 +1,7 @@
\section{Logical Algebra} \section{Logical Algebra}
\definition{} \definition{}
\textit{Logical operators} operate on the values $\{\text{True}, \text{False}\}$, \par \textit{Logical operators} operate on the values $\{\texttt{true}, \texttt{false}\}$, \par
just like algebraic operators operate on numbers. \par just like algebraic operators operate on numbers. \par
In this handout, we'll use the following operators: In this handout, we'll use the following operators:
\begin{itemize} \begin{itemize}
@ -9,7 +9,7 @@ In this handout, we'll use the following operators:
\item $\land$: and \item $\land$: and
\item $\lor$: or \item $\lor$: or
\item $\rightarrow$: implies \item $\rightarrow$: implies
\item $()$, parenthesis. \item $()$: parenthesis.
\end{itemize} \end{itemize}
The function of these is defined by \textit{truth tables}: The function of these is defined by \textit{truth tables}:
@ -19,10 +19,10 @@ The function of these is defined by \textit{truth tables}:
\hline \hline
$A$ & $B$ & $A \land B$ \\ $A$ & $B$ & $A \land B$ \\
\hline \hline
F & F & F \\ \texttt{F} & \texttt{F} & \texttt{F} \\
F & T & F \\ \texttt{F} & \texttt{T} & \texttt{F} \\
T & F & F \\ \texttt{T} & \texttt{F} & \texttt{F} \\
T & T & T \texttt{T}& \texttt{T} & \texttt{T}
\end{tabular} \end{tabular}
\hfill \hfill
\begin{tabular}{ c | c | c } \begin{tabular}{ c | c | c }
@ -30,10 +30,10 @@ The function of these is defined by \textit{truth tables}:
\hline \hline
$A$ & $B$ & $A \lor B$ \\ $A$ & $B$ & $A \lor B$ \\
\hline \hline
F & F & F \\ \texttt{F} & \texttt{F} & \texttt{F} \\
F & T & T \\ \texttt{F} & \texttt{T} & \texttt{T} \\
T & F & T \\ \texttt{T} & \texttt{F} & \texttt{T} \\
T & T & T \texttt{T} & \texttt{T} & \texttt{T}
\end{tabular} \end{tabular}
\hfill \hfill
\begin{tabular}{ c | c | c } \begin{tabular}{ c | c | c }
@ -41,10 +41,10 @@ The function of these is defined by \textit{truth tables}:
\hline \hline
$A$ & $B$ & $A \rightarrow B$ \\ $A$ & $B$ & $A \rightarrow B$ \\
\hline \hline
F & F & T \\ \texttt{F} & \texttt{F} & \texttt{T} \\
F & T & T \\ \texttt{F} & \texttt{T} & \texttt{T} \\
T & F & F \\ \texttt{T} & \texttt{F} & \texttt{F} \\
T & T & T \texttt{T} & \texttt{T} & \texttt{T}
\end{tabular} \end{tabular}
\hfill \hfill
\begin{tabular}{ c | c } \begin{tabular}{ c | c }
@ -52,8 +52,8 @@ The function of these is defined by \textit{truth tables}:
\hline \hline
$A$ & $\lnot A$ \\ $A$ & $\lnot A$ \\
\hline \hline
T & F \\ \texttt{T} & \texttt{F} \\
F & T \\ \texttt{F} & \texttt{T} \\
~ & ~ \\ ~ & ~ \\
~ & ~ \\ ~ & ~ \\
\end{tabular} \end{tabular}
@ -61,30 +61,40 @@ The function of these is defined by \textit{truth tables}:
\vspace{2mm} \vspace{2mm}
$A \land B$ is only true if both $A$ and $B$ are true. $A \lor B$ is true when $A$ or $B$ (or both) are true. \par $A \land B$ is \texttt{true} only if both $A$ and $B$ are \texttt{true}. $A \lor B$ is \texttt{true} if $A$ or $B$ (or both) are \texttt{true}. \par
$\lnot A$ is the opposite of $A$, which is why it looks like a \say{negative} sign. \par $\lnot A$ is the opposite of $A$, which is why it looks like a \say{negative} sign. \par
\vspace{2mm} \vspace{2mm}
$A \rightarrow B$ is a bit harder to understand. Read aloud, this is \say{$A$ implies $B$.} \par $A \rightarrow B$ is a bit harder to understand. Read aloud, this is \say{$A$ implies $B$.} \par
The only time $\rightarrow$ is false is when $T \rightarrow F$. This may seem counterintuitive, but it will make more sense as we progress through this handout. The only time $\rightarrow$ produces \texttt{false} is when $\texttt{true} \rightarrow \texttt{false}$.
This fact may seem counterintuitive, but will make more sense as we progress through this handout. \par
\hint{
Think about it---if event $\alpha$ implies $\beta$, it is impossible for $\alpha$ to occur without $\beta$. \par
This is the only impossibility. All other variants are valid.
}
\problem{} \problem{}
Evaluate the following. Evaluate the following.
\begin{itemize} \begin{itemize}
\item $\lnot T$ \item $\lnot \texttt{T}$
\item $F \lor T$ \item $\texttt{F} \lor \texttt{T}$
\item $T \land T$ \item $\texttt{T} \land \texttt{T}$
\item $(T \land F) \lor T$ \item $(\texttt{T} \land \texttt{F}) \lor \texttt{T}$
\item $(T \land F) \lor T$ \item $(\texttt{T} \land \texttt{F}) \lor \texttt{T}$
\item $(\lnot (F \lor \lnot T) ) \rightarrow T$ \item $(\lnot (\texttt{F} \lor \lnot \texttt{T}) ) \rightarrow \texttt{T}$
\item $(F \rightarrow T) \rightarrow (\lnot F \lor \lnot T)$ \item $(\texttt{F} \rightarrow \texttt{T}) \rightarrow (\lnot \texttt{F} \lor \lnot \texttt{T})$
\end{itemize} \end{itemize}
\vfill \vfill
\pagebreak \pagebreak
\begin{instructornote} \begin{instructornote}
After the class has done a few definable set problems, you can try to provide some intuition for $\rightarrow$ with the following example. We can also think of $[x \geq 0] \rightarrow b$ as follows:
if $x$ isn't the kind of object we care about, we evaluate true and
check the next one. If $x$ \textit{is} the kind of object we care about
and $b$ is false, we have a counterexample to $[x \geq 0] \rightarrow b$,
and thus $\texttt{T} \rightarrow \texttt{F}$ must be false.
\vspace{2mm} \vspace{2mm}
@ -96,17 +106,13 @@ Evaluate the following.
If $(\text{F} \rightarrow *)$ returned false, statements like the above would be hard to write. \par If $(\text{F} \rightarrow *)$ returned false, statements like the above would be hard to write. \par
If $x$ is negative, $\varphi$ doesn't care whether or not it has a root. In this case, $\text{F} \rightarrow *$ must be true to avoid making whole $\forall$ false. If $x$ is negative, $\varphi$ doesn't care whether or not it has a root. In this case, $\text{F} \rightarrow *$ must be true to avoid making whole $\forall$ false.
\vspace{2mm}
You can think of $[x \geq 0] \rightarrow b$ as a \say{sanity check} in a program: if $x$ isn't the kind of object we care about, return true and check the next one. If $x$ \textit{is} the kind of object we care about and $b$ is false, we have a counterexample to $[x \geq 0] \rightarrow b$, and thus $T \rightarrow F$ must be false.
\end{instructornote} \end{instructornote}
\problem{} \problem{}
Evaluate the following. Evaluate the following.
\begin{itemize} \begin{itemize}
\item $A \rightarrow T$ for any $A$ \item $A \rightarrow \texttt{T}$ for any $A$
\item $(\lnot (A \rightarrow B)) \rightarrow A$ for any $A, B$ \item $(\lnot (A \rightarrow B)) \rightarrow A$ for any $A, B$
\item $(A \rightarrow B) \rightarrow (\lnot B \rightarrow \lnot A)$ for any $A, B$ \item $(A \rightarrow B) \rightarrow (\lnot B \rightarrow \lnot A)$ for any $A, B$
\end{itemize} \end{itemize}
@ -115,13 +121,14 @@ Evaluate the following.
\problem{} \problem{}
Show that $\lnot (A \rightarrow \lnot B)$ is equivalent to $A \land B$. \par Show that $\lnot (A \rightarrow \lnot B)$ is equivalent to $A \land B$. \par
That is, show that these give the same result for the same $A$ and $B$. \par That is, show that these expressions always evaluate to the same value given
the same $A$ and $B$. \par
\hint{Use a truth table} \hint{Use a truth table}
\vfill \vfill
\problem{} \problem{}
Can you express $A \lor B$ using only $\lnot$, $\rightarrow$, and $()$? Write an expression equivalent to $A \lor B$ using only $\lnot$, $\rightarrow$, and $()$?
\begin{solution} \begin{solution}
$((\lnot A) \rightarrow B)$ $((\lnot A) \rightarrow B)$
@ -131,6 +138,6 @@ Can you express $A \lor B$ using only $\lnot$, $\rightarrow$, and $()$?
Note that both $\land$ and $\lor$ can be defined using the other logical symbols. \par Note that both $\land$ and $\lor$ can be defined using the other logical symbols. \par
The only logical symbols we \textit{need} are $\lnot$, $\rightarrow$, and $()$. \par The only logical symbols we \textit{need} are $\lnot$, $\rightarrow$, and $()$. \par
We include $\land$ and $\lor$ to simplify our logical expressions. We include $\land$ and $\lor$ to simplify our expressions.
\pagebreak \pagebreak

54
Advanced/Definable Sets/parts/1 structures.tex
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@ -9,27 +9,31 @@ A \textit{universe} is a set of meaningless objects. Here are a few examples:
\end{itemize} \end{itemize}
\definition{} \definition{}
A \textit{structure} consists of a universe $U$ and a set of symbols. \par A \textit{structure} consists of a universe and a set of \textit{symbols}. \par
A structure's symbols give meaning to the objects in its universe. A structure's symbols give meaning to the objects in its universe.
\vspace{2mm} \vspace{2mm}
Symbols come in three types: Symbols come in three types:
\begin{itemize} \begin{itemize}
\item Constant symbols, which let us specify specific elements of our universe. \par \item \textit{Constant symbols}, which let us specify specific elements of our universe. \par
Examples: $0, 1, \frac{1}{2}, \pi$ Examples: $0, 1, \frac{1}{2}, \pi$
\vspace{2mm} \vspace{2mm}
\item Function symbols, which let us navigate between elements of our universe. \par \item \textit{Function symbols}, which let us navigate between elements of our universe. \par
Examples: $+, \times, \sin{x}, \sqrt{x}$ Examples: $+, \times, \sin{x}, \sqrt{x}$ \par
\note{In this handout, symbols we usually call \say{operators} are also called functions. \par
The only difference between $a + b$ and $+(a, b)$ is notation.}
\vspace{2mm} \vspace{2mm}
\item Relation symbols, which let us compare elements of our universe. \par \item \textit{Relation symbols}, which let us compare elements of our universe. \par
Examples: $<, >, \leq, \geq$ \par Examples: $<, >, \leq, \geq$ \par
\vspace{2mm} \vspace{2mm}
\end{itemize} \end{itemize}
The equality check $=$ is \textbf{not} a relation symbol. It is included in every structure by default. The equality check $=$ is \textbf{not} a relation symbol. It is included in every structure by default. \par
By definition, $a = b$ is true if and only if $a$ and $b$ are the same element of our universe.
\vspace{3mm} \vspace{3mm}
@ -42,7 +46,7 @@ $$
\vspace{2mm} \vspace{2mm}
This is a structure with the universe $\mathbb{Z}$ that contains the following symbols: This is a structure over the universe $\mathbb{Z}$ that provides the following symbols:
\begin{itemize} \begin{itemize}
\item Constants: \tab $\{0, 1\}$ \item Constants: \tab $\{0, 1\}$
\item Functions: \tab $\{+, -\}$ \item Functions: \tab $\{+, -\}$
@ -51,24 +55,37 @@ This is a structure with the universe $\mathbb{Z}$ that contains the following s
\vspace{2mm} \vspace{2mm}
If we look at our set of constant symbols, we see that the only integers we can directly refer to in this structure are 0 and 1. If we want any others, we must define them using the tools this structure offers. If we look at our set of constant symbols, we see that the only integers
we can directly refer to in this structure are 0 and 1. If we want any
others, we must define them using the tools this structure offers.
\vspace{1mm} \vspace{2mm}
To \say{define} an element of a set, we need to write a sentence that is only true for that element. For example, if we want to define 2 in the structure above, we could use the sentence $\varphi(x) = [1 + 1 = x]$. \par To \say{define} an element of a set, we need to write a sentence that is only true for that element. \par
Clearly, this is only true when $x = 2$. For example, if we want to define 2 in the structure above, \par
we could use the sentence \say{$2$ is the $x$ that satisfies $[1 + 1 = x]$.} \par
This is a valid definition because $2$ is the \textbf{only} element of $\mathbb{Z}$ for which $[1 + 1 = x]$
evaluates to \texttt{true}.
\problem{} \problem{}
Define $-1$ in $\Bigl( \mathbb{Z} ~\big|~ \{0, 1, +, -, <\} \Bigr)$. Define $-1$ in $\Bigl( \mathbb{Z} ~\big|~ \{0, 1, +, -, <\} \Bigr)$.
\begin{solution}
The sentences \say{$x$ where $[x + 1 = 0]$} and \say{$x$ where $[0 - 1 = x]$} both work.
\end{solution}
\vfill \vfill
\pagebreak \pagebreak
Let us formalize what we found in the previous two problems. \par Let us formalize what we found in the previous two problems. \par
\definition{} \definition{Formulas}
A \textit{formula} in a structure $S$ is a well-formed string of constants, functions, and relations. \par A \textit{formula} in a structure $S$ is a well-formed string
of constants, functions, relations, \par and logical operators.
\vspace{2mm} \vspace{2mm}
@ -77,15 +94,20 @@ For the sake of time, I will not provide a formal definition. It isn't particula
\vspace{2mm} \vspace{2mm}
As a quick example, the formula $\phi(x) = [1 + 1 = x]$ evaluates to \texttt{true} when $x$ is 2 \par
and to \texttt{false} otherwise.
\definition{Free Variables}
A formula can contain one or more \textit{free variables.} These are denoted $\varphi{(a, b, ...)}$. \par A formula can contain one or more \textit{free variables.} These are denoted $\varphi{(a, b, ...)}$. \par
Formulas with free variables let us define \say{properties} that certain objects have. \par Formulas with free variables let us define \say{properties} that certain objects have. \par
For example, $x$ is a free variable in the formula $\varphi(x) = [x > 0]$. \par For example, $x$ is a free variable in the formula above. \par
$\varphi(3)$ is true and $\varphi(-3)$ is false. \par $\varphi(2)$ is \texttt{true} and $\varphi(-3)$ is \texttt{false}. \par
\vspace{2mm} \vspace{2mm}
This \say{free variable} notation is much like the function notation you are used to: \par This \say{free variable} notation is much like the function notation we are used to: \par
$\varphi(x) = [x > 0]$ is similar to $f(x) = x + 1$, since the values of $\varphi(x)$ and $f(x)$ depend on $x$. $\varphi(x) = [x > 0]$ is similar to $f(x) = x + 1$, since the values of $\varphi(x)$ and $f(x)$ depend on $x$.

18
Advanced/Definable Sets/parts/2 quantifiers.tex
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@ -4,12 +4,14 @@ Recall the logical symbols we introduced earlier: $(), \land, \lor, \lnot, \righ
We will now add two more: $\forall$ (for all) and $\exists$ (exists). We will now add two more: $\forall$ (for all) and $\exists$ (exists).
\definition{} \definition{}
$\forall$ and $\exists$ are \textit{quantifiers}. They allow us to make statements about arbitrary symbols. $\forall$ and $\exists$ are \textit{quantifiers}. They allow us to make statements about arbitrary symbols. \par
\hint{Quantifiers are aptly named: they tell us \textit{how many} symbols satisfy a certain sentence.}
\vspace{2mm} \vspace{2mm}
Let's look at $\forall$ first. Let $\varphi(x)$ be a formula. \par Let's look at $\forall$ first. If $\varphi(x)$ is a formula, \par
Then, the formula $\forall x ~ \varphi(x)$ says \say{$\varphi$ is true for all possible $x$.} the formula $\forall x ~ \varphi(x)$ is true only if $\varphi$ is true for all $x$ in our universe.
\vspace{1mm} \vspace{1mm}
@ -18,8 +20,8 @@ In English, this means \say{For any $x$, $x$ is bigger than zero,} or simply \sa
\vspace{3mm} \vspace{3mm}
$\exists$ is very similar: the formula $\exists x ~ \varphi(x)$ states that there is at least one $x$ that makes $\varphi$ true. \par $\exists$ is very similar: the formula $\exists x ~ \varphi(x)$ is true if ther is at least one $x$ for which $\varphi(x)$ is true. \par
For example, $\exists ~ (0 < x)$ means \say{there is a positive number in our set}. For example, $\exists ~ (0 < x)$ means \say{there is a positive number in our set.}
\vspace{4mm} \vspace{4mm}
@ -64,7 +66,7 @@ What's the difference between $\exists x ~ \forall y ~ (x \leq y)$ and $\forall
\hint{In $\mathbb{R}^+$, the first is false and the second is true. $\mathbb{R}^+$ does not contain zero.} \hint{In $\mathbb{R}^+$, the first is false and the second is true. $\mathbb{R}^+$ does not contain zero.}
\begin{solution} \begin{solution}
If $\exists x$ is inside $\forall y$, $x$ depends on $y$. We can have a different value of $x$ for every $y$. \par If $\exists x$ is inside $\forall y$, $x$ depends on $y$. We may pick a different value of $x$ for every $y$. \par
If $\exists x$ is outside, $x$ is fixed \textit{before} we check all $y$. If $\exists x$ is outside, $x$ is fixed \textit{before} we check all $y$.
\end{solution} \end{solution}
@ -98,10 +100,10 @@ Define $-1$ in $\Bigl( \mathbb{Z} ~\big|~ \{0, <\} \Bigr)$
\problem{} \problem{}
Let $\varphi(x)$ be a formula. \par Let $\varphi(x)$ be a formula. \par
Define $(\forall x ~ \varphi(x))$ using logical symbols and $\exists$. Write a formula equivalent to $[~ \forall x ~ \varphi(x) ~]$ using only logical symbols and $\exists$.
\begin{solution} \begin{solution}
$\Bigl(\forall x ~ \varphi(x)\Bigr)$ is true iff $\lnot \Bigl(\exists x ~ \lnot \varphi(x) \Bigr)$ is true. $[~ \forall x ~ \varphi(x) ~]$ is true if and only if $[~ \lnot \exists x ~ \lnot \varphi(x) ~]$ is true.
\end{solution} \end{solution}
\vfill \vfill

66
Advanced/Definable Sets/parts/3 sets.tex
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@ -3,37 +3,40 @@
Armed with $(), \land, \lor, \lnot, \rightarrow, \forall,$ and $\exists$, we have enough tools to define sets. Armed with $(), \land, \lor, \lnot, \rightarrow, \forall,$ and $\exists$, we have enough tools to define sets.
\definition{Set-Builder Notation} \definition{Set-Builder Notation}
Say we have a condition $c$. \par Say we have a sentence $\varphi(x)$. \par
The set of all elements that satisfy that condition can be written as follows: The set of all elements that satisfy that sentence can be written as follows:
$$ \begin{equation*}
\{ x ~|~ \text{$c$ is true} \} \{ x ~|~ \varphi(x) \}
$$ \end{equation*}
This is read \say{The set of $x$ where $c$ is true} or \say{The set of $x$ that satisfy $c$.} This is read \say{The set of $x$ where $\varphi$ is true} or \say{The set of $x$ that satisfy $\varphi$.}
\vspace{2mm} \vspace{2mm}
For example, take the formula $\varphi(x) = \exists y ~ (y + y = x)$. \par For example, take the formula $\varphi(x) = \exists y ~ (y + y = x)$. \par
The set of all even integers can then be written The set of all even integers can then be written as
$$ $$
\{ x ~|~ \varphi(x) \} = \{ x ~|~ \exists y ~ (y + y = x) \} \{ x ~|~ \exists y ~ (y + y = x) \}
$$ $$
\definition{Definable Sets} \definition{Definable Sets}
Let $S$ be a structure with a universe $U$. \par Let $S$ be a structure with a universe $U$. \par
We say a subset $M$ of $U$ is \textit{definable} if we can write a formula that is true for some $x$ iff $x \in M$. We say a subset $M$ of $U$ is \textit{definable} if we
can write a formula that is true for some $x$ iff $x \in M$.
\vspace{4mm} \vspace{4mm}
For example, consider the structure $\Bigl( \mathbb{Z} ~\big|~ \{+\} \Bigr)$ \par For example, consider the structure $\big\langle~ \mathbb{Z} ~\big|~ \{+\} ~\big\rangle$ \par
\vspace{2mm}
Only even numbers satisfy the formula $\varphi(x) = \exists y ~ (y + y = x)$, \par Only even numbers satisfy the formula $\varphi(x) = \exists y ~ (y + y = x)$, \par
So we can define \say{the set of even numbers} as $\{ x ~|~ \exists y ~ (y + y = x) \}$. \par so we can define \say{the set of even numbers} as $\{ x ~|~ \exists y ~ (y + y = x) \}$. \par
Remember---we can only use symbols that are available in our structure! Remember---we can only use symbols that are available in our structure!
\problem{} \problem{}
Is the empty set definable in any structure? When is the empty set definable?
\begin{solution}
Always: $\{ x ~|~ \lnot (x = x) \}$
\end{solution}
\vfill \vfill
@ -44,31 +47,22 @@ Define $\{0, 1\}$ in $\Bigl( \mathbb{Z}^+_0 ~\big|~ \{<\} \Bigr)$
\begin{instructornote} \begin{instructornote}
Here's an interesting fact: Here's an interesting fact:
A finite set of definable elements is always definable. Why? \par A finite set of definable elements is always definable. \note{(Why?)} \par
An infinite set of definable elements may not be definable. An infinite set of definable elements might not be definable.
\end{instructornote} \end{instructornote}
\vfill \vfill
\problem{} \problem{}
Define the set of prime numbers in $\Bigl( \mathbb{Z} ~\big|~ \{\times, \div, <\} \Bigr)$ Define the set of prime numbers in $\Bigl( \mathbb{Z} ~\big|~ \{\times, \div, <\} \Bigr)$
\vfill
\vfill
\pagebreak \pagebreak
\problem{}
Define the set of nonreal numbers in $\Bigl( \mathbb{C} ~\big|~ \{\text{real}(z)\} \Bigr)$ \par
\hint{$\text{real}(z)$ gives the real part of a complex number: $\text{real}(3 + 2i) = 3$}
\hint{$z$ is nonreal if $x \in \mathbb{C}$ and $x \notin \mathbb{R}$}
\begin{solution}
$\Bigl\{ x ~\bigl|~ \text{real}(x) \neq x \Bigr\}$
\end{solution}
\vfill
\problem{} \problem{}
@ -77,7 +71,7 @@ Define $\mathbb{R}^+_0$ in $\Bigl( \mathbb{R} ~\big|~ \{\times\} \Bigr)$ \par
\vfill \vfill
\problem{} \problem{}
Let $\bigtriangleup$ be a relational symbol. $a \bigtriangleup b$ holds iff $a$ divides $b$. \par Let $\bigtriangleup$ be a relational symbol. $a \bigtriangleup b$ is true if and only if $a$ divides $b$. \par
Define the set of prime numbers in $\Bigl( \mathbb{Z}^+ ~\big|~ \{ \bigtriangleup \} \Bigr)$ \par Define the set of prime numbers in $\Bigl( \mathbb{Z}^+ ~\big|~ \{ \bigtriangleup \} \Bigr)$ \par
\vfill \vfill
@ -93,7 +87,7 @@ Define $\mathbb{Z}^+_0$ in $\Bigl( \mathbb{Z} ~\big|~ \{\times, +\} \Bigr)$
\problem{} \problem{}
Define $<$ in $\Bigl( \mathbb{Z} ~\big|~ \{\times, +\} \Bigr)$ \par Define $<$ in $\Bigl( \mathbb{Z} ~\big|~ \{\times, +\} \Bigr)$ \par
\hint{We can't formally define a relation yet. Don't worry about that for now. \\ \hint{We can't formally define a relation yet. Don't worry about that for now. \\
You can repharase this question as \say{given $a,b \in \mathbb{Z}$, can you write a sentence that is true iff $a < b$?}} You can repharase this question as \say{given $a,b \in \mathbb{Z}$,\\*/ write a sentence that is only true if $a < b$?}}
\vfill \vfill
\pagebreak \pagebreak
@ -111,20 +105,24 @@ Define $\{-2, 2\}$ in $S$.
\vfill \vfill
\problem{} \problem{}
Let $P$ be the set of all subsets of $\mathbb{Z}^+_0$. This is called a \textit{power set}. \par Let $\mathcal{P}$ be the set of all subsets of $\mathbb{Z}^+_0$. This is called the \textit{power set} of $\mathbb{Z}^+_0$. \par
Let $S$ be the structure $( P ~|~ \{\subseteq\})$ \par Let $S$ be the structure $( \mathcal{P} ~|~ \{\subseteq\})$ \par
\problempart{} \problempart{}
Show that the empty set is definable in $S$. \par Show that the empty set is definable in $S$. \par
\hint{Defining $\{\}$ with $\{x ~|~ x \neq x\}$ is \textbf{not} what we need here. \\ \hint{Defining $\{\}$ with $\{x ~|~ x \neq x\}$ is \textbf{not} what we need here. \\
We need $\varnothing \in P$, the \say{empty set} element in the power set of $\mathbb{Z}^+_0$.} We need $\varnothing \in \mathcal{P}$, the \say{empty set} element in the power set of $\mathbb{Z}^+_0$.}
\vfill
\problempart{} \problempart{}
Let $x \Bumpeq y$ be a relation on $P$. $x \Bumpeq y$ holds if $x \cap y \neq \{\}$. \par Let $x \Bumpeq y$ be a relation on $\mathcal{P}$. $x \Bumpeq y$ holds if $x \cap y \neq \{\}$. \par
Show that $\Bumpeq$ is definable in $S$. Show that $\Bumpeq$ is definable in $S$.
\vfill
\problempart{} \problempart{}
Let $f$ be a function on $P$ defined by $f(x) = \mathbb{Z}^+_0 - x$. This is called the \textit{complement} of the set $x$. \par Let $f$ be the function on $\mathcal{P}$ defined by $f(x) = \mathbb{Z}^+_0 - x$. This is called the \textit{complement} of $x$. \par
Show that $f$ is definable in $S$. Show that $f$ is definable in $S$.
\vfill \vfill

2
Advanced/Definable Sets/parts/4 equivalence.tex
View File

@ -7,7 +7,7 @@ This is read \say{$S$ satisfies $\varphi$}
\definition{} \definition{}
Let $S$ and $T$ be structures. \par Let $S$ and $T$ be structures. \par
We say $S$ and $T$ are \textit{equivalent} and write $S \equiv T$ if for any formula $\varphi$, $S \models \varphi \Longleftrightarrow T \models \varphi$. \par We say $S$ and $T$ are \textit{equivalent} (and write $S \equiv T$) if for any formula $\varphi$, $S \models \varphi \Longleftrightarrow T \models \varphi$. \par
If $S$ and $T$ are not equivalent, we write $S \not\equiv T$. If $S$ and $T$ are not equivalent, we write $S \not\equiv T$.
\problem{} \problem{}