OpenLLM-France/wikimedia huggingface dataset:
# Consciousness Studies/Notes for the Contributors

Please ensure that your proposed contribution has not been covered elsewhere in the text. (The sections on what needs to be explained and phenomenal and access consciousness should certainly be read before making a contribution). The term materialist is used frequently in the text to describe someone who believes that nineteenth century science is adequate to explain the world (Whitehead's usage), the opposite of materialism is not spiritualism etc. - both modern physicalists and dualists are opposed to old fashioned materialism.
Please ensure that references and, preferably, links to original sources are provided for each assertion. These should be added to the "references" sections for each module.
An attempt has been made to give the book the sort of structure that might be expected in a science textbook (history, the empirical problem, experimental data, theories). The book is structured as follows:

## Introduction

* The problem of consciousness. What are all the problems that need to be explained when investigating consciousness? Specific theories of consciousness do not go here.
** Description of the empirical problem.
** The conflict. The debate about WHEN and WHERE of phenomenal consciousness and whether such a when and where could exist at all.
** Breaking down the philosophical and scientific problems. The when and where of the world. Demonstrating that the problems of consciousness have repercussions extending into the nature of space and time itself.
** The idea of machine consciousness. This is about the somewhat separate problem of whether machines could be conscious.
** Consciousness and the measurement problem. Another somewhat separate problem.
* Historical Review - placed here rather than after the introduction because it is a deeply academic review that might "put off" readers who have not had their interest kindled.

## NEUROSCIENCE

* Contemporary explanations.
* Other explanations. Yes, less widely known theories can go here, on separate pages or linked.
As can be seen from the structure of the book, this text is largely about the form of "consciousness" that is of interest to neuroscientists. Perhaps a second part of this book, classified under "philosophy" rather than "neuroscience" is required to explain the consequences of naive and direct realism in sociological and political theory and the consequences of idealism and dualism in religious thought.
The book is designed for undergraduate neuroscience and philosophy modules on consciousness.
# Consciousness Studies/Other Explanations

Consciousness-only is the foundation of a buddhist theory known as vijnanavada. Proponents, notably the Yogacara school, suggest that the sum of experience exists only in our minds. Philosophers recognition of this view as subjective idealism is a matter of discussion because these traditions often deny the existence of ontological subject. These views are rooted in the denial of existence of any kind of ontological substance (as Matter, Soul, God, etc.). Consciousness-only views can also be found in taoist philosophy, notably Lao Tzu and Chuang Tzu.
# Consciousness Studies/Models Of Access Consciousness

There have been numerous attempts to model reflex and access consciousness. These models, being connectionist and information systems based, do not model phenomenal consciousness but are essential steps in understanding global brain function.

## Neural networks

Neural networks achieve information processing by establishing connections between processing units in a system of processors that have similar characteristics. Neural networks are used for classifying data. The processing units serve the function of both filtering and storing information.
This is a stub and requires expansion

### Classification of sensory stimuli

The path from transducers to a single neuron that responds to a single complex stimulus.
This is a stub and requires expansion

### Classification of motor control

From premotor activity to skilled behaviour.
This is a stub and requires expansion

### Olfaction: classification out of chaos?

This is a stub and requires expansion

## Quantum information processing

This is a stub and needs expanding
# Consciousness Studies/Medieval Concepts

It is sometimes said that before the 17th century, there was no distinction between conscience and consciousness (both being referenced by the same Latin word 'conscius—from 'conscientia'). One feature of the soul being its ability to know itself, and the soul being witnessed also by God, there was a sense in which conscience was a knowledge of the inner life shared with God. In fact, Boris Hennig proposes that Descartes did not invent an entirely new sense of conscience, but if anything adapted this Augustinian sense of the term. The Augustinians themselves can be said to have adapted this use of conscius from the Latin Stoics like Seneca. On the other hand the Ancient Greek philosophers were clear that there was a problem of mind and the work of Descartes can be seen as more directly connected with Aristotle rather than with medieval philosophy.

## References

* ↑ "Cartesian Conscientia", British Journal for the History of Philosophy 15(3) 2007: 455 – 484.
# Circuit Theory/TF Examples/Example14

Find a partical solution to one of the re-occurring examples in this book. Given V$_S$, find i$_o$. This has to be split into two transfer functions:
    ${\frac {\mathbb {I} _{o}}{\mathbb {V} _{s}}}={\frac {\mathbb {I} _{o}}{\mathbb {I} _{t}}}{\frac {\mathbb {I} _{t}}{\mathbb {V} _{s}}}$
Which is of the form:
    ${\frac {\mathbb {I} _{o}}{\mathbb {V} _{s}}}={\text{current divider}}*{\text{admittance}}$
The current divider is going to be:
    ${\frac {\mathbb {I} _{o}}{\mathbb {I} _{t}}}={\frac {\frac {1}{R_{3}}}{{\frac {1}{R_{3}}}+{\frac {1}{sL_{2}+{\frac {1}{sC_{2}}}}}}}={\frac {sL_{2}+{\frac {1}{sC_{2}}}}{R_{3}+sL_{2}+{\frac {1}{sC_{2}}}}}$
Substituting into the transfer function:
    ${\frac {\mathbb {I} _{o}}{\mathbb {V} _{s}}}={\frac {sL_{2}+{\frac {1}{sC_{2}}}}{R_{3}+sL_{2}+{\frac {1}{sC_{2}}}}}*{\frac {1}{R_{1}+{\frac {1}{sC_{1}}}+sL_{1}+{\frac {1}{{\frac {1}{R_{3}}}+{\frac {1}{sL_{2}+{\frac {1}{sC_{2}}}}}}}}}$
Substituting and solving using this mupad code:
    ${\frac {\mathbb {I} _{o}}{\mathbb {V} _{s}}}={\frac {1000*(s^{2}+5000000.0)*s}{s^{4}+5000000.0*s^{3}+2.000015*10^{12}*s^{2}+3.5*10^{13}*s+5.0*10^{13}}}$
The differential equation for this would be:
    $1000({d^{2}V_{s} \over dt^{2}}+5000000.0){dV_{s} \over dt}={d^{4}I_{o} \over dt^{4}}+5000000.0{d^{3}I_{o} \over dt^{3}}+2.000015*10^{12}{d^{2}I_{o} \over dt^{2}}+3.5*10^{13}{dI_{o} \over dt}+5.0*10^{13}I_{o}$
Set the source V$_s$ to zero and then find I$_o$ by adding together Second order techniques (Real roots, are underdamped. Equal roots are critically damped and imaginary roots are overdamped.
Solving for the zeros (setting denominator to zero):
    ${\text{zeros}}=-4561551.036,-438431.4637,-15.93127461,-1.569297305$
So in this case the solution would be:
    $i_{o}=Ae^{-4561551.036t}+Be^{-438431.4637t}+Ce^{-15.93127461t}+De^{-1.569297305}+E$
The next step would be to find four initial conditions of i$_0$ (and final condition). Most texts don't attempt to solve this complex of a problem. The steady state particular solution was easily found long ago with phasors. Transfer functions help to find a general solution to any type of voltage source. And it helps find the homogeneous solution that describes a ramp up when the circuit is first turned on. What circuits do engineers design that is worth this effort? Filters. The fact that today most of this is done digitally just means that all this needs to be repeated in a digital context. (Simulation becomes reality!). So using math tools to solve these problems in preparation for understanding more advanced tools is reason for continuing this thread of content.
So on to filters (Bode Plots .. Fourier Analysis) where the constants disappear and we stay in the complex domain. We will need the poles for to analyze as a filter (setting numerator to zero):
    ${\text{poles}}=0,-2236.067977i,2236.067977i$
# Circuit Theory/TF Examples/Example33

Find i$_o$(t) if V$_s$(t) = 1 + cos(3t).

## Choose Starting Point

Because of the initial conditions, going to start with V$_c$(t) and then work our way through the initial conditions to i$_o$.

## Transfer Function

    $H(s)={\frac {V_{c}}{V_{s}}}={\frac {\frac {1}{sC}}{{\frac {1}{sC}}+{\frac {1}{{\frac {1}{R_{1}}}+{\frac {1}{sL}}}}+R_{2}}}$
The MuPad commands are going to be:
Which results in:
    $H(s)={\frac {8s+4}{8s^{2}+11s+4}}$

## Homogeneous Solution

Set the denominator of the transfer function to 0 and solve for s:
Imaginary roots:
    $s_{1,2}={\frac {-11\pm {\sqrt {7}}i}{16}}$
So the solution has the form:
    $V_{c_{h}}=e^{-{\frac {11t}{16}}}(A\cos {\frac {7t}{16}}+B\sin {\frac {7t}{16}})+C$

## Particular Solution

After a very long time the capacitor opens, no current flows, so all the source drop is across the capacitor. The source is a unit step function thus:
    $V_{c_{p}}=1$

## Initial Conditions

Adding the particular and homogenous solutions, get:
    $V_{c}(t)=1+e^{-{\frac {11t}{16}}}(A\cos {\frac {7t}{16}}+B\sin {\frac {7t}{16}})+C$
Doing the final condition again, get:
    $V_{c}(\infty )=1=1+C\Rightarrow C=0$
Which implies that C is zero.
From the given initial conditions, know that V$_c$(0+) = 0.5 so can find A:
    $V_{c}(0_{+})=0.5=1+A\Rightarrow A=-0.5$
Finding B is more difficult. From capacitor terminal relation:
The total current is:
    $i_{T}(t)=C{dV_{c} \over dt}={\frac {11e^{-{\frac {11t}{16}}}}{16}}(0.5\cos {\frac {7t}{16}}-B\sin {\frac {7t}{16}})+{\frac {7e^{-{\frac {11t}{16}}}}{16}}(0.5\sin {\frac {7t}{16}}+B\cos {\frac {7t}{16}})$
The loop equation can be solved for the voltage across the LR parallel combination:
    $V_{C}+V_{LR}+R_{2}C{dV_{c} \over dt}-V_{s}=0$
    $V_{LR}=V_{s}-V_{C}-R_{2}i_{t}=1-V_{c}-1.5*i_{t}$
We know from the inductor terminal relation that:
    $i_{L}={\frac {1}{L}}\int V_{LR}dt+C_{1}$
At this point mupad gave up and went numeric. In any case, it is clear from t = ∞ where the inductor current has to be zero that the integration constant is zero. This enables us to compute B from the inductor initial condition.
Set the time to zero, set I$_L$ equal to the initial condition of .2 amps and solve for B:
And get that B is -0.2008928571 ...
The desired answer is i$_o$ which is just V$_{LR}$/R_1. To calculate need to start new mupad session because t is zero now. Start with:
Repeat the above commands up to VLR and then add:
$i_{o}=3e^{-{\frac {11t}{16}}}(0.0879\cos {\frac {7t}{16}}-0.219\sin {\frac {7t}{16}})-0.0625e^{-{\frac {11t}{16}}}(0.5\cos {\frac {7t}{16}}+0.201\sin {\frac {7t}{16}})$
# Circuit Theory/TF Examples/Example34/Vc
## Starting Point

Starting point of i$_L$ looks good. Can get the voltage across the inductor and the parallel components. Then a voltage divider is going to help get to the V$_c$ initial condition.

## Transfer Function

    $H(s)={\frac {i_{o}}{i_{s}}}={\frac {\frac {1}{R_{2}+{\frac {1}{sC}}}}{{\frac {1}{sL}}+{\frac {1}{R_{1}}}+{\frac {1}{R_{2}+{\frac {1}{sC}}}}}}$
    ${\frac {i_{o}}{i_{s}}}={\frac {2s^{2}}{8s^{2}+11s+4}}$

## Homogeneous Solution

Set the denominator of the transfer function to 0 and solve for s:
    $s_{1,2}={\frac {-11\pm {\sqrt {7}}i}{16}}$
So the solution is going to have the form:
    $i_{L_{h}}=e^{\frac {11t}{16}}(A\cos {\frac {7}{16}}+B\sin {\frac {7}{16}})+C$

## Particular Solution

After a very long time the inductor shorts, all the current flows through it so:
    $i_{o}=0$

## Initial Conditions

Adding the particular and homogeneous solutions, get:
    $i_{o}(t)=e^{-{\frac {11t}{16}}}(A\cos {\frac {7t}{16}}+B\sin {\frac {7t}{16}})+C$
Doing the final condition again, get:
    $i_{o}(\infty )=0=C\Rightarrow C=0$
Let's try for V$_c$ first. From the terminal relation for a capacitor:
    $V_{c}(t)={\frac {1}{C}}\int i_{o}dt$
We know that initially V$_c$ = 0.5 so at t=0 can find equation for A and B:
At this point mupad goes numeric and get this equation:
    $A=-0.6363636364B-0.4829545455$
Need another equation. Can find V$_t$ by adding V$_r$ and V$_c$. Then from V$_t$ can find expression for the current through the inductor and visit it's initial condition.
The integration constant is going to be zero because after a long time V$_C$ is zero (the inductor shorts).
Mupad goes numeric.
At this point have to figure out the integration constant. After a long time, the inductor's current is going to be 1 because it shorts the current source. Looking at IL in the mupad window can see that every term is multiplied by e$^{-0.6875t}$ which is going to zero as t goes to ∞. This means the integration constant is 1.
So add 1 to IL, then set t=0 and I$_L$ = 0.5 and again solve for A and B:
Get this equation:
    $A=0.4107683001B+0.5463551119$
Now need to solve the two equations and two unknowns:
Get:
    $A=0.142578125,B=-0.9829799107$
So now have time domain expression for i$_o$ step response:
    $i_{o}(t)\mu (t)=e^{-{\frac {11t}{16}}}(0.143\cos {\frac {7t}{16}}-0.983\sin {\frac {7t}{16}})$

## Impulse Response

The impulse response is the derivative of the step response:

## Convolution Integral

The first step is to substitute into i_s for t:
Now form the convolution integral:
Replacing y with t:
There is going to be an integration constant. This value can not be determined because the driving function oscillates. The initial conditions of the inductor and capacitor have already been visited. More information (like a specific value at a future time) is needed in order to compute an integration constant.
Thus the final answer is:
    $i_{o}=0.344cos(2t)+0.314sin(2t)-0.201e^{-0.6875t}cos(0.438t)+1.29e^{-0.6875t}sin(0.438t)-0.143$
# Circuit Theory/TF Examples/Example34/iL
## Starting Point

Starting point of i$_L$ looks good. Can get the voltage across the inductor and the parallel components. Then a voltage divider is going to help get to the V$_c$ initial condition.

## Transfer Function

    $H(s)={\frac {i_{o}}{i_{s}}}={\frac {\frac {1}{R_{2}+{\frac {1}{sC}}}}{{\frac {1}{sL}}+{\frac {1}{R_{1}}}+{\frac {1}{R_{2}+{\frac {1}{sC}}}}}}$
    ${\frac {i_{o}}{i_{s}}}={\frac {2s^{2}}{8s^{2}+11s+4}}$

## Homogeneous Solution

Set the denominator of the transfer function to 0 and solve for s:
    $s_{1,2}={\frac {-11\pm {\sqrt {7}}i}{16}}$
So the solution is going to have the form:
    $i_{L_{h}}=e^{\frac {11t}{16}}(A\cos {\frac {7}{16}}+B\sin {\frac {7}{16}})+C$

## Particular Solution

After a very long time the inductor shorts, all the current flows through it so:
    $i_{o}=0$

## Initial Conditions

Adding the particular and homogeneous solutions, get:
    $i_{o}(t)=e^{-{\frac {11t}{16}}}(A\cos {\frac {7t}{16}}+B\sin {\frac {7t}{16}})+C$
Doing the final condition again, get:
    $i_{o}(\infty )=0=C\Rightarrow C=0$
Let's try for V$_c$ first. From the terminal relation for a capacitor:
    $V_{c}(t)={\frac {1}{C}}\int i_{o}dt$
We know that initially V$_c$ = 0.5 so at t=0 can find equation for A and B:
At this point mupad goes numeric and get this equation:
    $A=-0.6363636364B-0.4829545455$
Need another equation. Can find V$_t$ by adding V$_r$ and V$_c$. Then from V$_t$ can find expression for the current through the inductor and visit it's initial condition.
The integration constant is going to be zero because after a long time V$_C$ is zero (the inductor shorts).
Mupad goes numeric.
At this point have to figure out the integration constant. After a long time, the inductor's current is going to be 1 because it shorts the current source. Looking at IL in the mupad window can see that every term is multiplied by e$^{-0.6875t}$ which is going to zero as t goes to ∞. This means the integration constant is 1.
So add 1 to IL, then set t=0 and I$_L$ = 0.5 and again solve for A and B:
Get this equation:
    $A=0.4107683001B+0.5463551119$
Now need to solve the two equations and two unknowns:
Get:
    $A=0.142578125,B=-0.9829799107$
So now have time domain expression for i$_o$ step response:
    $i_{o}(t)\mu (t)=e^{-{\frac {11t}{16}}}(0.143\cos {\frac {7t}{16}}-0.983\sin {\frac {7t}{16}})$

## Impulse Response

The impulse response is the derivative of the step response:

## Convolution Integral

The first step is to substitute into i_s for t:
Now form the convolution integral:
Replacing y with t:
There is going to be an integration constant. This value can not be determined because the driving function oscillates. The initial conditions of the inductor and capacitor have already been visited. More information (like a specific value at a future time) is needed in order to compute an integration constant.
Thus the final answer is:
    $i_{o}=0.344cos(2t)+0.314sin(2t)-0.201e^{-0.6875t}cos(0.438t)+1.29e^{-0.6875t}sin(0.438t)-0.143$
# Circuit Theory/TR

A terminal is a place where a wire comes out of an electrical component.
Terminal relations for the 2 wire devices studied so far (resistors, capacitors and inductors), describe the voltage across the two terminals and the current entering and leaving the terminals (which is always the same).
The terminal relations for a resistor are $v=i*R$ (called ohms law)
The terminal relations for a capacitor are $i=C*dv/dt$
The terminal relations for an inductor are $v=L*di/dt$
A power supply doesn't have a terminal relationship or equation relating current and voltage. Either the current is constant (current supply) or voltage is constant (voltage supply). The power is varied depending upon what the circuit asks for.
One of the best places to build up your intuition or build your own animated circuit simulations is at falstad
# Circuit Theory/TF Examples

These are some worked problems:
    2 resistor, 2 inductor, 2 capacitor problem
    series RLC circuit with 2 initial conditions
    parallel RLC circuit with 2 initial conditions
# Circuit Theory/TF Examples/Example34

$i_{s}=1+3cos(2t)$, find i$_o$
This problem is done three different ways, reflecting three different transfer function starting points which imply three different paths to visit the initial conditions. Click on the transfer functions below:
    $H(s)={\frac {i_{0}}{i_{s}}}$
    $H(s)={\frac {i_{L}}{i_{s}}}$
    $H(s)={\frac {V_{c}}{i_{s}}}$
# Annotated King Reference Manual/Program Structure
## Compilation Units
### Rationale

At library level, subprogram_body shall be only procedure.
# Annotated Republic of China Case Law/Judicial Yuan Interpretation No. 176

| ←Judicial Yuan Interpretation No. 175 | Judicial Yuan Interpretation No. 176 | Judicial Yuan Interpretation No. 177→ |

## Date

13 August 1982

## Issue

The criminality to circulate forged document abroad? The Control Yuan would petition for interpretation in Taiwan.

## Holding

In the Criminal Code, the crimes of Article 216 listed in Clause 5 of Article 5 include circulating documents of Article 213, not Article 210, 212, or 215.

### Dissent

* Justice Shih-Ron CHEN (part)

## Effect

This interpretation would specify the applicability of Article 216 of the Criminal Code of the Republic of China abroad.

## References

* Translation by WELLINGTON L. KOO, Formosa Transnational attorney at law
* Annotated Republic of China Laws/Criminal Code of the Republic of China, Articles 5, 210, 212, 215, 216 (1969-12-26)
# Annotated Republic of China Case Law/Judicial Yuan Interpretation No. 22

| ←Judicial Yuan Interpretation No. 21 | Judicial Yuan Interpretation No. 22 | Judicial Yuan Interpretation No. 23→ |

## Date

4 August 1953

## Issue

Are Members of the Legislative Yuan and the Control Yuan paid public officials? The Ministry of National Defense requested the Executive Yuan to petition the Judicial Yuan for constitutional interpretation.

## Holding

Members of the Legislative Yuan and the Control Yuan are paid public officials.

## Effect

Just keeping the status quo.

## References

* Translation by Li-Chih Lin, Esq., J.D.
* Constitution of the Republic of China, Articles 62, 103 (1947-12-25)
* Annotated Republic of China Laws/Temporary Act on Legislators’ and Ombudsmen’s Annual Pay and Pay for Duty-related Expenses
# Annotated King Reference Manual/Statements

This page is work in progress.

## Case Statements
### Examples
### Rationale

- Normal case statements may appear in any sequence of statements. Exception-handling case statements may only appear in an exception handler.
The exception_identifier must be the identifier following when of the enclosing exception handler.

### Discussion
## Wait statements
### Examples
### Rationale

The wait of statement is for a relative amount of time, and blocks the task for the number of seconds given. It can be read as, "Perform a wait of this many seconds." The wait for statement takes a Duration value and blocks the task until that time arrives, unless it is in the past. It can be read as, "Wait for this time to arrive." The duration is the number of seconds since the epoch, which is not defined, but meaningful values can be constructed using the standard library.

### Discussion
## Select Statements
### Rationale

A blocking call would be a call to any subprogram of a passive task.
# Annotated Republic of China Case Law/Judicial Yuan Interpretation No. 2

| ←Judicial Yuan Interpretation No. 1 | Judicial Yuan Interpretation No. 2 | Judicial Yuan Interpretation No. 3→ |

## Date

6 January 1949

## Issue

Shall uniform interpretation be mentioned when a governmental agency petitions for it?

## Holding

Uniform interpretation shall be mentioned when a governmental agency petitions for it.

## Effect

When a governmental agency may petition for unified interpretation is clearer.

## References

* Translation by Cheng-Yi HUANG
* Constitution of the Republic of China, Article 78 and 173 (1947-12-25)
# Annotated Republic of China Case Law/Judicial Yuan Interpretation No. 1

| Judicial Yuan Interpretation No. 1 | Judicial Yuan Interpretation No. 2→ |

## Date

6 January 1949

## Issue

A Member of the Legislative Yuan still remains so upon entering a government post? Because of a Legislator entering a government post, official petitions would be sent to the Judicial Yuan for constitutional interpretation.

## Holding

No Member of the Legislative Yuan shall concurrently hold a government post per Article 75 of the Constitution of the Republic of China. Entering a government post shall resign as a Member of the Legislative Yuan. Entering a government post without resignation clearly shows discontinuing the Membership of the Legislative Yuan and shall be deemed to resign upon entering a government post.

## Effect

Hereafter the membership of the Legislative Yuan will be automatically lost upon entering a government post. This is similar to certain countries automatically removing citizenship and nationality upon voluntarily becoming foreign citizens and nationals.

## References

* Translation by Robert Huai-Ching Tsai
* Constitution of the Republic of China, Article 75 (1947-12-25)
# Annotated Republic of China Case Law/Judicial Yuan Interpretation No. 161

| ←Judicial Yuan Interpretation No. 160 | Judicial Yuan Interpretation No. 161 | Judicial Yuan Interpretation No. 162→ |

## Date

18 January 1980

## Issue

Calculating the effective date of a law regulation would include the enactment or publication date? Having different internal opinions in the Government of the Republic of China, The Executive Yuan would petition for unified interpretation in Taiwan.

## Holding

Calculating the effective date of a law regulation per Article 13 of the Central Regulation Standard Act shall include the enactment or publication date.

### Dissent

* Justice Shih-Ron CHEN
* Justice Rei-Kuang YIAU
* Justice Yu-Po CHENG

## Effect

Hereafter a Taiwanese law or regulation stipulated as to be enforced from the date of enactment or publication means that if enacted or published today, it will become effective the day after tomorrow.

## References

* Translation by J. P. Fa
* Annotated Republic of China Laws/Central Regulation Standard Act#Chapter 3 The Enforcement of the Regulation, Article 13 (1970-08-31)