avewright/proofwiki-math huggingface dataset:
Theorem: Distribution of Macroscopic Electric Charge within Conductor

Let $B$ be a body made out of an electrically conducting substance.
Then it is not possible for different volume elements of $B$ to have different (macroscopic) electric fields caused by intrinsic imbalance of the electric charges distributed throughout $B$.

Proof:
In an electrical conductor, electric charges can flow, because of the action of conduction electrons within the substance.
Hence an electric charge in one point within $B$ will flow to another point within $B$ under a potential difference caused by differences in (macroscopic) electric fields within $B$.
Hence any imbalances caused by internal (macroscopic) electric fields will not be possible.
{{qed}}
Theorem: T3 Space with Sigma-Locally Finite Basis is T4 Space

Let $T = \struct {S, \tau}$ be a $T_3$ topological space.
Let $\BB$ be a $\sigma$-locally finite basis.
Then:
:$T$ is a $T_4$ space

Proof:
Let $A$ and $B$ be disjoint closed subsets of $T$.
=== Lemma 1 ===
{{:T3 Space with Sigma-Locally Finite Basis is T4 Space/Lemma 1}}{{qed|lemma}}
From {{Lemma|T3 Space with Sigma-Locally Finite Basis is T4 Space|1}}:
{{begin-itemize}}
{{item||there exists a countable open cover $\UU {{=}} \set {U_n : n \in \N}$ of $A$:
{{begin-itemize}}
{{item||$\forall n \in \N : U_n^- \cap B {{=}} \O$}}
{{end-itemize}} }}
{{item||there exists a countable open cover $\VV {{=}} \set {V_n : n \in \N}$ of $B$:
{{begin-itemize}}
{{item||$\forall n \in \N : V_n^- \cap A {{=}} \O$}}
{{end-itemize}} }}
{{end-itemize}}
From Countable Open Covers Condition for Separated Sets:
:$A$ and $B$ can be separated in $T$ 
By definition of separated:
:there exists $U, V \in \tau$ such that $A \subseteq U, B \subseteq V$ and $U \cap V = \O$
Since $A$ and $B$ were arbitrary disjoint closed subsets of $T$, by definition, $T$ is a $T_4$ space.
{{qed}}
Theorem: T3 Space with Sigma-Locally Finite Basis is T4 Space

Let $T = \struct {S, \tau}$ be a $T_3$ topological space.
Let $\BB$ be a $\sigma$-locally finite basis.
Then:
:$T$ is a $T_4$ space

Proof:
From T3 Space with Sigma-Locally Finite Basis is Paracompact:
:$T$ is a paracompact space.
From $T_3$ Space is Fully $T_4$ iff Paracompact:
:$T$ is a fully $T_4$ space.
From Fully $T_4$ Space is $T_4$:
:$T$ is a $T_4$ space.
{{qed}}
Theorem: Characterization of T3 Space

Let $T = \struct {S, \tau}$ be a topological space. 
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$T$ is a $T_3$ space}}
{{item|(2):|$\forall F \subseteq S : S \setminus F \in \tau, y \in S \setminus F : \exists V \in \tau : y \in V, V^- \cap F {{=}} \O$}}
{{item|(3):|$\forall U \in \tau, y \in U : \exists V \in \tau : y \in V, V^- \subseteq U$}}
{{end-itemize}}
where $V^-$ denotes the closure of $V$ in $T$

Proof:
=== Statement $(1)$ implies Statement $(2)$ ===
Let $T$ be a $T_3$ space.
Let $F \subseteq S : S \setminus F \in \tau$.
Let $y \in S \setminus F$.
By definition of $T_3$ space:
:$\exists V, W \in \tau : y \in V, F \subseteq W : V \cap W = \O$
From Subset of Set Difference iff Disjoint Set:
:$V \subseteq S \setminus W$
By definition of closed set:
:$S \setminus W$ is closed in $T$
From Closure of Subset of Closed Set of Topological Space is Subset:
:$V^- \subseteq S \setminus W$
From Set Difference with Subset is Superset of Set Difference:
:$S \setminus W \subseteq S \setminus F$
From Subset Relation is Transitive:
:$V^- \subseteq S \setminus F$
From Subset of Set Difference iff Disjoint Set:
:$V^- \cap F = \O$
The result follows.
{{qed|lemma}}
Theorem: Open Star Convex Set Minus Countable Set is Path-Connected in Real Euclidean Space

Let $k \ge 2$ be a natural number.
Let $A \subset \R^k$ be a star convex set, with star center $a \in A$.
Let $G \subset \R^k$ be finite or countable.
Suppose $a \notin G$.
Then $A \setminus G$ is path-connected.

Proof:
Let $x \in A \setminus G$ be an arbitrary point.
By definition of dimension, there is some vector $\vec v \in \R^k$ that is linearly independent from $\vec {x a}$.
Now, consider the mapping $f: \R \to \R^k$:
:$\map {\vec f} k = \vec v + k \vec {x a}$
The vector $\vec {x a}$ is linearly independent from every vector in the image of $\vec f$.
:For, $\paren {\vec v + k \vec {x a} } - k \vec {x a} = \vec v$, which is linearly independent from $\vec {x a}$.
Likewise, every pair of vectors in $\Img f$ is linearly independent.
:For, $\dfrac 1 {k_1 - k_2} \paren {\paren {\vec v + k_1 \vec {x a} } - \paren {\vec v + k_2 \vec {x a} } } = \vec {x a}$
:If either $k_1$ or $k_2$ is $0$, then we are done.
:Otherwise, $\dfrac {k_2} {k_2 - k_1} \paren {\paren {\vec v + k_1 \vec {x a} } - \dfrac {k_1} {k_2} \paren {\vec v + k_2 \vec {x a} } } = \vec v$.
Associate with each $g \in G$ the unique $k \in \R$, if it exists, such that:
:$\exists t \in \R: g = x + t \map {\vec f} k$
That is, the real number $k$ such that $g$ lines on the line extending from $x$ in the direction $\map {\vec f} k$.
By Real Numbers are Uncountably Infinite, $\size \R > \size G$.
Therefore, by Surjection iff Cardinal Inequality, the association is not a surjection.
Thus, there is some $k_0 \in \R$ such that no $g \in G$ lies on the line extending from $x$ in the direction $\map {\vec f} {k_0}$.
By definition of open set, there exists some open ball $\map {B_\epsilon} x \subset A$.
Define the mapping $h: \openint 0 \epsilon \to \R^k$:
:$\map h m = x + m \dfrac {\map {\vec f} {k_0} } {\size {\map {\vec f} {k_0} } }$
In words, map each real number in the interval to the point that is distance $m$ in the direction of $\map {\vec f} {k_0}$ from $x$.
As $\vec {x a}$ and $\map {\vec f} {k_0}$ are linearly independent, it follows that the vectors:
:$\overrightarrow {\map h m a}$
are are pairwise linearly independent.
We can again associate the $g \in G$ with the unique $m$ such that $g$ lies on the line $\map h m + t \overrightarrow {\map h m a}$.
Each of those lines lies in $A$, by definition of star convex.
As Real Interval is Uncountably Infinite, it follows that there is some $m_0$ for which no $g$ lies on the corresponding line.
But then, the concatenation of the lines from $x$ to $\map h {m_0}$, and then from $\map h {m_0}$ to $a$ does not intersect any $g \in G$.
Therefore, the same path exists in $A \setminus G$.
Finally, for any $y, z \in A \setminus G$, the concatenation of the path from $y$ to $a$, followed by the reverse of $z$ to $a$, forms a path from $y$ to $z$.
Therefore, $A \setminus G$ is path-connected.
{{qed}}
Category:Real Euclidean Spaces
Category:Path-Connected Sets
t6k89kgqwfoz1ws4uo2ddvdwccl0xik
Theorem: Macroscopic Electric Field in Conductor in Static Equilibrium is Zero

Let $B$ be a body made out of an electrically conducting substance.
Let $B$ be in static equilibrium.
Then the macroscopic electric field within $B$ is everywhere zero.

Proof:
By definition of static equilibrium there are no forces upon $B$.
That includes forces caused by an electric field.
From Distribution of Macroscopic Electric Charge within Conductor, the macroscopic electric field is constant throughout $B$.
An electric field arises through differences in electric charges.
As we have seen, there are no such differences.
Hence the result.
{{qed}}
Theorem: T3 Space with Sigma-Locally Finite Basis is Paracompact

Let $T = \struct {S, \tau}$ be a $T_3$ topological space.
Let $\BB$ be a $\sigma$-locally finite basis of $T$.
Then $T$ is a paracompact.

Proof:
Let $\UU$ be an open cover of $T$.
Let $\VV = \set {B \in \BB : \exists U \in \UU : B \subseteq U}$.
Then:
:$\VV \subseteq \BB$
From Subset of Sigma-Locally Finite Set of Subsets is Sigma-Locally Finite:
:$\VV$ is $\sigma$-locally finite
Let $x \in S$.
By definition of open cover:
:$\exists U \in \UU : x \in U$
By definition of basis:
:$\exists B \in \BB : x \in B \subseteq U$
Hence:
:$B \in \VV$
It follows that $\VV$ is an open cover by definition.
By definition, $\VV$ is an open refinement of $\UU$.
It has been shown that:
:every open cover of $T$ has an open $\sigma$-locally finite refinement.
From Characterization of Paracompactness in $T_3$ Space:
:$T$ is paracompact.
{{qed}}
Category:T3 Spaces
Category:Sigma-Locally Finite Bases
Category:Paracompact Spaces
7ak9y4ipqp5zmhubgoigxfww3njminr
Theorem: Nonexistence of Complex Matrices whose Commutator equals Identity

Let $d \in \N_{> 0}$ be a positive natural number.
Let $\mathbf A, \mathbf B \in \C^{d \times d}$ be complex matrices.
{{explain|Same applies to real matrices too, yes? I don't have the source to support that, but I don't see why not.<br>This theorem includes the result on the real case, because real matrices are especially complex matrices.<br/>Obvious though it is, the real case might be worth adding as a corollary}}
Let $\mathbf I$ be the $d \times d$ identity matrix.
Then there is no $\mathbf A$, $\mathbf B$ such that $\mathbf A \mathbf B - \mathbf B \mathbf A = \mathbf I$.

Proof:
We have that complex numbers form a commutative ring.
{{AimForCont}} there are $\mathbf A$, $\mathbf B$ such that $\mathbf A \mathbf B - \mathbf B \mathbf A = \mathbf I$.
Then:
{{begin-eqn}}
{{eqn | l = \map \tr {\mathbf I} 
      | r = d
      | c = {{Defof|Trace of Matrix}}
}}
{{eqn | r = \map \tr {\mathbf A \mathbf B - \mathbf B \mathbf A}
}}
{{eqn | r = \map \tr {\mathbf A \mathbf B} - \map \tr {\mathbf B \mathbf A}
      | c = Trace of Sum of Matrices is Sum of Traces
}}
{{eqn | r = 0
      | c = Trace of Product of Matrices
}}
{{end-eqn}}
This is a contradiction.
{{qed}}
Theorem: Cumulative Distribution Function of Logistic Distribution

Let $X$ be a continuous random variable with the logistic distribution.
Then the cumulative distribution function of $X$ is:
:$\map {F_X} x = \dfrac 1 {1 + \map \exp {- \dfrac {x - \mu} s} }$

Proof:
The derivative of $F_X$ is:
{{begin-eqn}}
{{eqn | l = \map {F'_X} x
      | r = \paren {\frac 1 {1 + \map \exp {- \frac {x - \mu} s} } }'
}}
{{eqn | r = \frac {\paren 1' \paren {1 + \map \exp {- \frac {x - \mu} s} } - \paren 1 \paren {1 + \map \exp {- \frac {x - \mu} s} }'} {\paren {1 + \map \exp {- \frac {x - \mu} s} }^2}
      | c = Quotient Rule for Derivatives
}}
{{eqn | r = \frac {0 - \paren {\map \exp {-\frac {\paren {x - \mu} } s} }'} {\paren {1 + \map \exp {- \frac {x - \mu} s} }^2}
      | c = Derivative of Constant, Sum Rule for Derivatives
}}
{{eqn | r = \frac {- \map {\exp'} {-\frac {\paren {x - \mu} } s} \paren {-\frac {\paren {x - \mu} } s}'} {\paren {1 + \map \exp {- \frac {x - \mu} s} }^2}
      | c = Chain Rule for Derivatives
}}
{{eqn | r = \frac {\map \exp {- \frac {x - \mu} s} \paren {\frac x s - \frac \mu s}'} {\paren {1 + \map \exp {- \frac {x - \mu} s} }^2}
      | c = Derivative of Exponential Function, Derivative of Constant Multiple
}}
{{eqn | r = \frac {\map \exp {- \frac {x - \mu} s} } {s\paren {1 + \map \exp {- \frac {x - \mu} s} }^2}
      | c = Derivative of Constant Multiple, Derivative of Identity Function, Derivative of Constant
}}
{{eqn | r = \map {f_X} x
      | c = {{Defof|Logistic Distribution}}
}}
{{end-eqn}}
By the Fundamental Theorem of Calculus:
:$\ds \int_a^b \map {f_X} x \rd x = \bigintlimits {\map {F_X} x} {x \mathop = a} {x \mathop = b}$
Therefore:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^x \map {f_X} \lambda \rd \lambda
      | r = \lim_{a \to -\infty} \int_a^x \map {f_X} \lambda \rd \lambda
      | c = {{Defof|Improper Integral}}
}}
{{eqn | r = \lim_{a \to -\infty} \paren {\map {F_X} x - \map {F_X} a}
      | c = Result above
}}
{{eqn | r = \lim_{a \to -\infty} \map {F_X} x - \lim_{a \to -\infty} \map {F_X} a
      | c = Sum Rule for Limits of Real Functions
}}
{{eqn | r = \map {F_X} x - \lim_{a \to -\infty} \frac 1 {1 + \map \exp {- \frac {a - \mu} s} }
      | c = Limit of Constant Function
}}
{{eqn | r = \map {F_X} x - \lim_{b \to \infty} \frac 1 {1 + \map \exp b}
      | c = Limit to Infinity of Linear Function, where $b = - \frac {a - \mu} s$
}}
{{eqn | r = \map {F_X} x - \lim_{c \to \infty} \frac 1 c
      | c = Exponential Tends to Zero and Infinity, where $c = 1 + \map \exp b$
}}
{{eqn | r = \map {F_X} x
      | c = Limit to Infinity of Reciprocal Function
}}
{{end-eqn}}
Hence the result.
{{qed}}
Category:Logistic Distribution
f1fhwtqvjbmbjskd8j0u6pdgccbylqj
Theorem: Union of Turing Machines

Let $T_1, T_2$ be Turing machines.
Let $\Sigma_1, \Sigma_2$ be the input symbols of $T_1, T_2$, respectively.
Let $L_1, L_2$ be the languages accepted by $T_1, T_2$, respectively.
There exists a Turing machine $T$ such that:
* The input symbols of $T$ are $\Sigma_1 \cup \Sigma_2$.
* The language accepted by $T$ is $L_1 \cup L_2$.
* $T$ halts on some input {{iff}} it is in the accepted language, or both $T_1$ and $T_2$ halt on it.

Proof:
Let:
:$T_1 = \tuple {Q_1, \Sigma_1, \Gamma_1, \delta_1, q_1, B, F_1}$
:$T_2 = \tuple {Q_2, \Sigma_2, \Gamma_2, \delta_2, q_2, B, F_2}$
As implied by the notation, the blank symbols of the two machines will be identified with each other, and denoted as $B$.
Construct the $2$-tape Turing machine $T' = \tuple{Q, \Sigma, \Gamma, \delta, q_0, B, F}$ as follows:
:$Q = \set {q_0, q_L} \cup \paren {Q_1 \cup \set {q_H} } \times \paren {Q_2 \cup \set {q_H} }$
:$\Sigma = \Sigma_1 \cup \Sigma_2$
:$\Gamma = \Gamma_1 \cup \Gamma_2$
:$F = F_1 \times Q_2 \cup Q_1 \times F_2$
:$\map \delta {q, s, t} = \begin{cases}
\tuple {q_0, s, s, R, R} & : q = q_0 \land s \ne B \\
\tuple {q_L, B, B, L, L} & : q = q_0 \land s = B \\
\tuple {q_L, s, t, L, L} & : q = q_L \land s \ne B \\
\tuple {\tuple {q_1, q_2}, B, B, R, R} & : q = q_L \land s = B \\
\tuple {\tuple {a', b'}, s', t', d_1, d_2} & : q = \tuple {a, b}
\end{cases}$
::where $\tuple {a', s', d_1} = \map {\delta_1} {a, s}$ if $\map {\delta_1} {a, s}$ is defined, and $\tuple {a, s, S}$ otherwise; and likewise for $\delta_2$.
The following special cases will apply to $\delta$:
* $\map {\delta} {\tuple {q_H, q_H}, s, t}$ is undefined for any $s, t$
* $\map {\delta} {\tuple {a, b}, s, t}$ is undefined if at least one of $a \in F_1$ or $b \in F_2$ is true.
The state $q_0$ copies the input string from tape $1$ to tape $2$, until it encounters a blank.
State $q_L$ moves both tapes back to the start of the input string.
After this, $T_1$ and $T_2$ are represented on the two tapes of $T'$, in the natural way.
$T'$ will accept and halt when either machine enters one of its accepting states, as required.
When either machine halts, that is when $\map {\delta_1} {a, s}$ is undefined, $T'$ changes the respective state to $q_H$, and does not alter its tape or state after that.
If both machines halt without accepting, $T'$ does the same.
The result follows from Multitape Turing Machine Reduces to Turing Machine.
{{qed}}
Category:Turing Machines
28k3wt6strnlbhlgac39ptd8u8mu2as
Theorem: Finite State Machine is Turing Computable

Let $F = \tuple {S, A, I, \Sigma, T}$ be a finite state machine.
Then there exists a Turing machine $T$ that:
* Has input language $\Sigma$.
* Accepts exactly the same language as $F$.
* Halts on every input.

Proof:
Define the Turing machine:
:$T = \tuple {S \cup \set {H}, \Sigma, \Sigma \cup \set B, \delta, I, B, \set H}$
where:
:$\map \delta {s, \sigma} = \tuple {\map T {s, \sigma}, \sigma, R}$ if $q \in S$ and $\sigma \ne B$
:$\map \delta {s, B} = \tuple {H, B, R}$ if $s \in A$
The machine behaves identically to $F$ while the input is being read.
When there is no more input, if $\sigma \in A$, then $T$ transitions to the $H$ state, which is the accepting state.
If $\sigma \notin A$, then there is no transition defined, so $T$ halts without accepting.
As $T$ only moves $R$, it will eventually reach the end of the input word, and one of these two conditions must hold.
{{qed}}
Category:Turing Machines
tcud4ifsosuwj6i0m1h1sdo1ur2t435
Theorem: Conjunction of Finite State Machines

Let $\Sigma$ be a finite set.
Let $F_1 = \tuple {S_1, A_1, I_1, \Sigma, T_1}$ and $F_2 = \tuple {S_2, A_2, I_2, \Sigma, T_2}$ be finite state machines.
Then there exists a finite state machine $F$ such that $\map L F = \map L {F_1} \cap \map L {F_2}$.

Proof:
Construct $F = \tuple {S_1 \times S_2, A_1 \times A_2, \tuple {I_1, I_2}, \Sigma, T}$ where:
:$\map T {s_1, s_2, \Sigma} = \tuple {\map {T_1} {s_1}, \map {T_2} {s_2} }$
The machines are simultaneously emulated in the first and second components of the state tuple.
When the input word is exhausted, $F$ accepts {{iff}} both $F_1$ and $F_2$ are in an accepting state, as required.
{{qed}}
Category:Abstract Machines
iddxl2g91c6pkr7zrbnyim0ogfy3pk1
Theorem: Disjunction of Finite State Machines

Let $\Sigma$ be a finite set.
Let $F_1 = \tuple {S_1, A_1, I_1, \Sigma, T_1}$ and $F_2 = \tuple {S_2, A_2, I_2, \Sigma, T_2}$ be finite state machines.
Then there exists a finite state machine $F$ such that $\map L F = \map L {F_1} \cup \map L {F_2}$.

Proof:
Construct $F = \tuple {S_1 \times S_2, A_1 \times S_2 \cup S_1 \times A_2, \tuple {I_1, I_2}, \Sigma, T}$ where:
:$\map T {s_1, s_2, \Sigma} = \tuple {\map {T_1} {s_1}, \map {T_2} {s_2} }$
The machines are simultaneously emulated in the first and second components of the state tuple.
When the input word is exhausted, $F$ accepts {{iff}} at least one of $F_1$ and $F_2$ is in an accepting state, as required.
{{qed}}
Category:Abstract Machines
tfrcosbf5o96xzy38v2yl6slluq3n1j
Theorem: Negation of Finite State Machine

Let $F = \tuple {S, A, I, \Sigma, T}$ be a finite state machine.
Then there exists a finite state machine $F'$ such that $\map L {F'} = \map \complement {\map L F}$.

Proof:
Construct $F' = \tuple {S, S \setminus A, I, \Sigma, T}$.
The state transitions are identical to those of $F$.
However, when the input word is exhausted, the final state is an accepting state {{iff}} it is not an accepting state in $F$, as required.
{{qed}}
Category:Abstract Machines
fr1mwzhwds5tq73o9vka8feontorjrm
Theorem: Symbol Count by Finite State Machine

Let $\Sigma$ be a finite set.
Let $\Pi \subseteq \Sigma$.
For a finite sequence $w = \sequence {a_1, a_2, \dotsc, a_n}$ of elements of $\Sigma$:
:Let $\map c w = \size {\set {i : a_i \in \Pi} }$
Let $k \in \N$.
Then, for each of the following, there exists a finite state machine whose accepted languages is all $w \in \Sigma^*$ such that the condition holds:
* $\map c w < k$
* $\map c w \le k$
* $\map c w > k$
* $\map c w \ge k$
* $\map c w = k$
* $\map c w \ne k$

Proof:
=== $\map c w < k$ ===
Define $F = \tuple {S, A, I, \Sigma, T}$ where:
:$S = \set {s_0, s_1, \dotsc, s_k}$
:$A = \set {s_0, s_1, \dotsc, s_{k - 1} }$
:$I = s_0$
:$\map T {s_i, \sigma} = \begin{cases}
s_{i + 1} & : \sigma \in \Pi \land i < k \\
s_i & : \text {otherwise}
\end{cases}$
The machine being in the state $s_i$ indicates that $i$ symbols in $\Pi$ have been read so far, up to a maximum of $k$.
So long as the word ends before the $k$-th symbol in $\Pi$ is read, the machine accepts.
If a $k$-th symbol is read, it is no longer possible for the machine to transition out of $s_k$, and the machine rejects.
{{qed|lemma}}
Theorem: Intersection of Turing Machines

Let $T_1, T_2$ be Turing machines.
Let $\Sigma_1, \Sigma_2$ be the input symbols of $T_1, T_2$, respectively.
Let $L_1, L_2$ be the languages accepted by $T_1, T_2$, respectively.
There exists a Turing machine $T$ such that:
* The input symbols of $T$ are $\Sigma_1 \cap \Sigma_2$.
* The language accepted by $T$ is $L_1 \cap L_2$.
* $T$ halts on some input {{iff}} it is in the accepted language, or either $T_1$ or $T_2$ halt on it without accepting.

Proof:
Let:
:$T_1 = \tuple {Q_1, \Sigma_1, \Gamma_1, \delta_1, q_1, B, F_1}$
:$T_2 = \tuple {Q_2, \Sigma_2, \Gamma_2, \delta_2, q_2, B, F_2}$
As implied by the notation, the blank symbols of the two machines will be identified with each other, and denoted as $B$.
Construct the $2$-tape Turing machine $T' = \tuple{Q, \Sigma, \Gamma, \delta, q_0, B, F}$ as follows:
:$Q = \set {q_0, q_L} \cup Q_1 \times Q_2$
:$\Sigma = \Sigma_1 \cap \Sigma_2$
:$\Gamma = \Gamma_1 \cup \Gamma_2$
:$F = F_1 \times F_2$
:$\map \delta {q, s, t} = \begin{cases}
\tuple {q_0, s, s, R, R} & : q = q_0 \land s \ne B \\
\tuple {q_L, B, B, L, L} & : q = q_0 \land s = B \\
\tuple {q_L, s, t, L, L} & : q = q_L \land s \ne B \\
\tuple {\tuple {q_1, q_2}, B, B, R, R} & : q = q_L \land s = B \\
\tuple {\tuple {a', b'}, s', t', d_1, d_2} & : q = \tuple {a, b}
\end{cases}$
::where:
:::$\tuple {a', s', d_1} = \map {\delta_1} {a, s}$ if $\map {\delta_1} {a, s}$ is defined and $a \notin F_1$
:::$\tuple {a', s', S} = \tuple {a, s, S}$ if $a \in F_1$
::and likewise for $\delta_2$
The following special case also applies to $\delta$:
:$\map \delta {q, s, t}$ is undefined if $q \in F$
::That is, if $q = \tuple {a, b}$ such that $a \in F_1$ and $b \in F_2$
The states $q_0$ and $q_L$ effectively copy the input string onto tape $2$ without altering tape $1$.
The two machines are represented on the separate tapes in the natural way.
Once $T_1$ or $T_2$ enters an accepting state, emulation stops and it is saved in that state indefinitely.
Once the second machine enters an accepting state as well, the machine $T$ accepts and halts, by the special case.
If either machine halts without accepting, the transition function $\delta$ is undefined for that state, and $T$ halts.
{{qed}}
Category:Turing Machines
tnxfn1zv0lnb0wj835wlh9a3veio14n
Theorem: Real and Imaginary Parts of Holomorphic Function are Harmonic

Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.
Let $f: D \to \C$ be a holomorphic complex function on $D$.
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be the two real-valued functions defined as:
{{begin-eqn}}
{{eqn | l = \map u {x, y}
      | r = \map \Re {\map f {x + i y} }
}}
{{eqn | l = \map v {x, y}
      | r = \map \Im {\map f {x + i y} }
}}
{{end-eqn}}
Then $u$ and $v$ are harmonic functions.

Proof:
By Cauchy-Riemann Equations, $u$ and $v$ satisfy:
{{begin-eqn}}
{{eqn | n = 1
      | l = \dfrac {\partial u} {\partial x}
      | r = \dfrac {\partial v} {\partial y}
}}
{{eqn | n = 2
      | l = \dfrac {\partial u} {\partial y}
      | r = -\dfrac {\partial v} {\partial x}
}}
{{end-eqn}}
{{Explain|Need to justify some facts about continuity and differentiability for this argument to hold.}}
Consider the partial derivative of $(1)$ {{WRT|Differentiation}} $x$, and the partial derivative of $(2)$ {{WRT|Differentiation}} $y$:
{{begin-eqn}}
{{eqn | n = 3
      | l = \dfrac {\partial^2 u} {\partial x^2}
      | r = \dfrac {\partial^2 v} {\partial x \partial y}
}}
{{eqn | n = 4
      | l = \dfrac {\partial^2 u} {\partial y^2}
      | r = -\dfrac {\partial^2 v} {\partial y \partial x}
}}
{{end-eqn}}
By Clairaut's Theorem:
:$\dfrac {\partial^2 v} {\partial x \partial y} = \dfrac {\partial^2 v} {\partial y \partial x}$
Thus $(3) + (4)$ yields:
:$\dfrac {\partial^2 u} {\partial x^2} + \dfrac {\partial^2 u} {\partial y^2} = 0$
So $u$ is a harmonic function by definition.
{{qed|lemma}}
Now consider the partial derivative of $(1)$ {{WRT|Differentiation}} $y$, and the partial derivative of $(2)$ {{WRT|Differentiation}} $x$:
{{begin-eqn}}
{{eqn | n = 5
      | l = \dfrac {\partial^2 u} {\partial x \partial y}
      | r = \dfrac {\partial^2 v} {\partial y^2}
}}
{{eqn | n = 6
      | l = \dfrac {\partial^2 u} {\partial y \partial x}
      | r = -\dfrac {\partial^2 v} {\partial x^2}
}}
{{end-eqn}}
By Clairaut's Theorem :
:$\dfrac {\partial^2 u} {\partial x \partial y} = \dfrac {\partial^2 u} {\partial y \partial x}$
Thus $(5) - (6)$ yields:
:$0 = \dfrac {\partial^2 v} {\partial y^2} + \dfrac {\partial^2 v} {\partial x^2}$
So $v$ is a harmonic function by definition.
{{qed}}