nyuuzyou/wikis huggingface dataset:
D'Alembert's Formula

== Theorem ==
Let $t$ be time.
Let $x$ be position.
Let $\tuple {t, x} \stackrel u {\longrightarrow} \map u {t, x}: \R^2 \to \R$ be a twice-differentiable function in both variables. 
Let $x \stackrel \phi {\longrightarrow} \map \phi x: \R \to \R$ be a differentiable function. 
Let $x \stackrel \psi {\longrightarrow} \map \psi x: \R \to \R$ be a Riemann integrable function.
Let $c \in \R_{> 0}$ be a constant.
Then the solution to the partial differential equation:
:$\dfrac {\partial^2} {\partial t^2} \map u {x, t} = c^2 \dfrac {\partial^2} {\partial x^2} \map u {x, t}$
with initial conditions:
is given by:
:$\ds \map u {x, t} = \dfrac 1 2 \paren {\map \phi {x + c t} + \map \phi {x - c t} } + \dfrac 1 {2 c} \int_{x - c t}^{x + c t} \map \psi s \rd s$
The above solution formula is called d'Alembert's formula.
== Proof ==
The general solution to the $1$-D wave equation:
:$\dfrac {\partial^2} {\partial t^2} \map u {x, t} = c^2 \dfrac {\partial^2} {\partial x^2} \map u {x, t}$
is given by:
:$\map u {x, t} = \map f {x + c t} + \map g {x - c t}$
where $f, g$ are arbitrary twice-differentiable functions.
From initial conditions we have:
So we have:
Solving the equations give:
Integrating both equations and using Fundamental Theorem of Calculus:
for some constants $A, B$.
From $\map \phi x = \map f x + \map g x$, we have $A + B = 0$. 
Therefore:
== Historical Note ==
== Sources ==
* : Chapter $2$: Waves and Diffusions
* : Chapter $\S 2.1$: Continuous and linear maps. Linear transformations
Category:Partial Differential Equations
Dandelin's Theorem

== Theorem ==
Let $\CC$ be a double napped right circular cone with apex $O$.
Let $\PP$ be a plane which intersects $\CC$ such that:
:$\PP$ does not pass through $O$
:$\PP$ is not perpendicular to the axis of $\CC$.
Let $\EE$ be the conic section arising as the intersection between $\PP$ and $\CC$.
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
=== Foci ===
=== Directrices ===
</onlyinclude>
== Also known as ==
Dandelin's theorem is also seen referred to as Dandelin's construction.
== Also see ==
* Definition:Dandelin Spheres
Category:Dandelin's Theorem
Category:Dandelin Spheres
Category:Conic Sections
Dandelin's Theorem/Directrices

== Theorem ==
Let $\CC$ be a double napped right circular cone with apex $O$.
Let $\PP$ be a plane which intersects $\CC$ such that:
:$\PP$ does not pass through $O$
:$\PP$ is not perpendicular to the axis of $\CC$.
Let $\EE$ be the conic section arising as the intersection between $\PP$ and $\CC$.
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Let $\KK$ and $\KK'$ be the planes in which the ring-contacts of $\CC$ with $\SS$ and $\SS'$ are embedded respectively.
:The intersections of $\KK$ and $\KK'$ with $\PP$ form the directrices of $\EE$.
== Proof ==
== Also see ==
* Definition:Dandelin Spheres
== Sources ==
* : Chapter $\text {IV}$. The Ellipse: $1 \text a$. Focal properties
Category:Dandelin's Theorem
Dandelin's Theorem/Directrices/Proof

== Theorem ==
Let $\CC$ be a double napped right circular cone with apex $O$.
Let $\PP$ be a plane which intersects $\CC$ such that:
:$\PP$ does not pass through $O$
:$\PP$ is not perpendicular to the axis of $\CC$.
Let $\EE$ be the conic section arising as the intersection between $\PP$ and $\CC$.
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Then:
== Proof ==
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Let $P$ be a point on $\EE$.
Let $F$ and $F'$ be the points at which $\SS$ and $\SS'$ are tangent to $\PP$ respectively.
Let the generatrix of $\CC$ which passes through $P$ touch $\SS$ and $\SS'$ at $E$ and $E'$ respectively.
Let $\theta$ be half the opening angle of $\CC$.
Let $\phi$ be the inclination of $\PP$ to the axis of $\CC$.
Let $\PP$ intersect $\KK$ in the straight line $NX$.
Let $PN$ be constructed perpendicular to $NX$.
Let $PK$ be constructed perpendicular to $\KK$.
Then:
:$PK = PN \cos \phi$
Also:
:$PK = PE \cos \theta = PF \cos \theta$
Hence:
:$\dfrac {PF} {PN} = \dfrac {\cos \phi} {\cos \theta}$
which is constant.
Hence $NX$ is a directrix of $\EE$ by definition.
A similar argument and construction applies with respect to $\SS'$ and $\KK$.
 
== Theorem ==
* : Chapter $\text {IV}$. The Ellipse: $1 \text a$. Focal properties
Category:Dandelin's Theorem
Dandelin's Theorem/Foci

== Theorem ==
Let $\CC$ be a double napped right circular cone with apex $O$.
Let $\PP$ be a plane which intersects $\CC$ such that:
:$\PP$ does not pass through $O$
:$\PP$ is not perpendicular to the axis of $\CC$.
Let $\EE$ be the conic section arising as the intersection between $\PP$ and $\CC$.
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Then:
:$\SS$ and $\SS'$ are tangent to $\PP$ at the foci of $\EE$.
== Proof ==
== Also see ==
* Definition:Dandelin Spheres
== Sources ==
* : Chapter $\text {IV}$. The Ellipse: $1 \text a$. Focal properties
Category:Dandelin's Theorem
Dandelin's Theorem/Foci/Proof

== Theorem ==
Let $\CC$ be a double napped right circular cone with apex $O$.
Let $\PP$ be a plane which intersects $\CC$ such that:
:$\PP$ does not pass through $O$
:$\PP$ is not perpendicular to the axis of $\CC$.
Let $\EE$ be the conic section arising as the intersection between $\PP$ and $\CC$.
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Then:
== Proof ==
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Let $P$ be a point on $\EE$.
Let $F$ and $F'$ be the points at which $\SS$ and $\SS'$ are tangent to $\PP$ respectively.
Let the generatrix of $\CC$ which passes through $P$ touch $\SS$ and $\SS'$ at $E$ and $E'$ respectively.
Because $PF$ and $PE$ are both tangent to $\SS$:
:$PF = PE$
and similarly:
:$PF' = PE'$
=== Proof for Ellipse ===
Let $\EE$ be an ellipse.
:
Then $E$ and $E'$ are on the same side of $O$, and so:
:$PF + PF' = PE + PE'$
But $PE + PE'$ is a constant.
Hence $\EE$ fulfils the equidistance property with respect to the points $F$ and $F'$.
Hence $F$ and $F'$ are the foci of $\EE$.
=== Proof for Hyperbola ===
Let $\EE$ be a hyperbola.
:
Then $E$ and $E'$ are on the opposites sides of $O$, and so:
:$PF - PF' = PE - PE'$
But $PE - PE'$ is a constant.
Hence $\EE$ fulfils the equidistance property with respect to the points $F$ and $F'$.
Hence $F$ and $F'$ are the foci of $\EE$.
=== Proof for Parabola ===
Let $\EE$ be a parabola.
:
Because $\PP$ is parallel to the generatrix of $\CC$, it follows that:
:$\triangle PHE = \triangle PHE'$
and so:
:$PF = PE'$
and it is seen that this is the focus-directrix property of the parabola.
 
== Theorem ==
* : Chapter $\text {IV}$. The Ellipse: $1 \text a$. Focal properties
Category:Dandelin's Theorem
Darboux's Theorem

== Theorem ==
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Let $\ds \int_a^b \map f x \rd x$ be the definite integral of $\map f x$ over $\closedint a b$.
Then:
:$\ds m \paren {b - a} \le \int_a^b \map f x \rd x \le M \paren {b - a}$
where:
:$M$ is the maximum of $f$
:$m$ is the minimum of $f$
on $\closedint a b$.
=== Corollary ===
== Proof ==
This follows directly from the definition of definite integral:
From Continuous Image of Closed Interval is Closed Interval it follows that $m$ and $M$ both exist.
The closed interval $\closedint a b$ is a finite subdivision of itself.
By definition, the upper Darboux sum is $M \paren {b - a}$, and the lower Darboux sum is $m \paren {b - a}$.
The result follows.
== Sources ==
* : $\S 13.4$
* 
* 
* 
* 
Category:Darboux's Theorem
Category:Integral Calculus
Darboux's Theorem/Corollary

== Corollary to Darboux's Theorem ==
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Suppose that $\forall t \in \closedint a b: \size {\map f t} < \kappa$.
Then:
:$\ds \forall \xi, x \in \closedint a b: \size {\int_x^\xi \map f t \rd t} < \kappa \size {x - \xi}$
== Proof ==
Follows directly from Darboux's Theorem.
== Sources ==
* : $\S 13.6$
Category:Darboux's Theorem
Data Coding/Examples/Arbitrary Example 1

== Examples of Data Coding ==
Let us wish to calculate the mean and standard deviation of the numbers:
:$179 \cdotp 385$
:$179 \cdotp 387$
:$179 \cdotp 392$
We may multiply by $1000$ and then subtract $179 \, 387$ (which is the same as adding $-179 \, 387$) to give:
:$-2$
:$0$
:$5$
The mean is:
:$\dfrac {-2 + 0 + 5} 3 = 1$
The standard deviation is:
:$\sqrt {\dfrac {\paren {-2 - 1}^2 + \paren {0 - 1}^2 + {5 - 1}^2} 3} = \sqrt {\dfrac {26} 3} = 2 \cdotp 94$
Adding $179 \, 387$ to $1$ and multiplying by $\dfrac 1 {1000}$ gives the mean as:
:$179 \cdotp 388$
Multiplying $2 \cdotp 94$ by $\dfrac 1 {1000}$ gives the standard deviation as:
:$0 \cdotp 002 \, 94$
== Sources ==
* 
* 
Category:Examples of Data Coding
De Moivre's Formula

== Theorem ==
Let $z \in \C$ be a complex number expressed in complex form:
:$z = r \paren {\cos x + i \sin x}$
Then:
:$\forall \omega \in \C: \paren {r \paren {\cos x + i \sin x} }^\omega = r^\omega \paren {\map \cos {\omega x} + i \, \map \sin {\omega x} }$
=== Exponential Form ===
De Moivre's Formula can also be expressed thus in exponential form:
=== Integer Index ===
This result is often given for integer index only:
=== Rational Index ===
Some sources give it for rational index:
== Proof 1 ==
== Also presented as ==
== Also known as ==
== Historical Note ==
== Sources ==
* : $\S 6$: Complex Numbers: De Moivre's Theorem: $6.9$
* 
* 
* : $\S 4$: Complex Numbers: De Moivre's Theorem: $4.10.$
* 
* : Appendix $6$: Basic Algebra
Category:De Moivre's Formula
Category:Complex Analysis
De Moivre's Formula/Also known as

== De Moivre's Formula: Also known as == 
De Moivre's Formula is also known as De Moivre's Theorem.
== Sources ==
* : $\S 6$: Complex Numbers: De Moivre's Theorem: $6.9$
* 
* 
* : $\S 4$: Complex Numbers: De Moivre's Theorem: $4.10.$
* 
* : Appendix $6$: Basic Algebra
Category:De Moivre's Formula
De Moivre's Formula/Also presented as

== De Moivre's Formula: Also presented as == 
De Moivre's Formula is also often presented in the simpler form:
:$\forall \omega \in \C: \paren {\cos x + i \sin x}^\omega = \map \cos {\omega x} + i \, \map \sin {\omega x}$
== Sources ==
* 
* 
* 
Category:De Moivre's Formula
De Moivre's Formula/Exponential Form

== Theorem ==
:$\forall \omega \in \C: \paren {r e^{i \theta} }^\omega = r^\omega e^{i \omega \theta}$
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Powers: $3.7.16$
* : $\S 7$: Operations with Complex Numbers in Polar Form: $7.27$
* : $1$: Complex Numbers: Euler's Formula
Category:De Moivre's Formula
De Moivre's Formula/Integer Index

== Theorem ==
Let $z \in \C$ be a complex number expressed in complex form:
:$z = r \paren {\cos x + i \sin x}$
Then:
=== Positive Index ===
=== Negative Index ===
=== Corollary ===
== Also known as ==
De Moivre's Theorem.
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Powers: $3.7.17$
* : Appendix: Elementary set and number theory
* : $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $20$
Category:De Moivre's Formula
De Moivre's Formula/Integer Index/Corollary

== Corollary to De Moivre's Formula: Positive Integer Index ==
:$\forall n \in \Z: \paren {\cos x + i \sin x}^n = \map \cos {n x} + i \map \sin {n x}$
== Proof ==
$\cos x + i \sin x$ is a complex number expressed in polar form $\polar {r, \theta}$ whose complex modulus is $1$ and whose argument is $x$.
From De Moivre's Formula: Integer Index:
:$\forall n \in \Z: \paren {r \paren {\cos x + i \sin x} }^n = r^n \paren {\map \cos {n x} + i \map \sin {n x} }$
The result follows by setting $r = 1$.
== Sources ==
* : $\S 2$. Geometrical Representations
* : $1$ The complex plane: Complex numbers $\S 1.2$ The algebraic structure of the complex numbers
Category:De Moivre's Formula
De Moivre's Formula/Negative Integer Index

== Theorem ==
Let $z \in \C$ be a complex number expressed in complex form:
:$z = r \paren {\cos x + i \sin x}$
Then:
:$\forall n \in \Z_{\le 0}: \paren {r \paren {\cos x + i \sin x} }^n = r^n \paren {\map \cos {n x} + i \map \sin {n x} }$
== Proof ==
Let $n = 0$.
Then:
Now let $n \in \Z_{<0}$.
Let $n = -m$ where $m > 0$.
Thus:
== Sources ==
* : $\S 2$. Geometrical Representations
* : $1$: Complex Numbers: Supplementary Problems: De Moivre's Theorem: $94 \ \text{(a)}$
Category:De Moivre's Formula
De Moivre's Formula/Positive Integer Index

== Theorem ==
Let $z \in \C$ be a complex number expressed in polar form:
:$z = r \paren {\cos x + i \sin x}$
Then:
:$\forall n \in \Z_{>0}: \paren {r \paren {\cos x + i \sin x} }^n = r^n \paren {\map \cos {n x} + i \map \sin {n x} }$
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Also known as ==
De Moivre's Theorem.
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 3$. Geometric Representation of Complex Numbers
* : $\text {(iii)}$
Category:De Moivre's Formula
De Moivre's Formula/Positive Integer Index/Corollary

== Corollary to De Moivre's Formula: Positive Integer Index ==
:$\forall n \in \Z_{>0}: \paren {\cos x + i \sin x}^n = \map \cos {n x} + i \map \sin {n x}$
== Proof 1 ==
== Proof 2 ==
Category:De Moivre's Formula
De Moivre's Formula/Positive Integer Index/Corollary/Proof 1

== Corollary to De Moivre's Formula: Positive Integer Index ==
== Proof ==
$\cos x + i \sin x$ is a complex number expressed in polar form $\left\langle{r, \theta}\right\rangle$ whose complex modulus is $1$ and whose argument is $x$.
From De Moivre's Formula: Positive Integer Index:
:$\forall n \in \Z_{>0}: \left({r \left({\cos x + i \sin x}\right)}\right)^n = r^n \left({\cos \left({n x}\right) + i \sin \left({n x}\right)}\right)$
The result follows by setting $r = 1$.
== Sources ==
* : $\S 2$. Geometrical Representations: $(2.16)$
Category:De Moivre's Formula
De Moivre's Formula/Positive Integer Index/Proof 1

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\paren {r \paren {\cos x + i \sin x} }^n = r^n \paren {\map \cos {n x} + i \, \map \sin {n x} }$
$\map P 1$ is the case:
:$\paren {r \paren {\cos x + i \sin x} }^1 = r^1 \paren {\map \cos {1 x} + i \, \map \sin {1 x} }$
which is trivially true.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\paren {r \paren {\cos x + i \sin x} }^2 = r^2 \paren {\map \cos {n x} + i \, \map \sin {2 x} }$
From Product of Complex Numbers in Polar Form, we have:
:$r_1 \paren {\cos x_1 + i \sin x_1 } r_2 \paren {\cos x_2 + i \sin x_2} = r_1 r_2 \paren {\map \cos {x_1 + x_2} + i \, \map \sin {x_1 + x_2} }$
Setting $r_1 = r_2 = r$ and $x_1 = x_2 = x$ gives the result.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\paren {r \paren {\cos x + i \sin x} }^k = r^k \paren {\map \cos {k x} + i \, \map \sin {k x} }$
Then we need to show:
:$\paren {r \paren {\cos x + i \sin x} }^{k + 1} = r^{k + 1} \paren {\map \cos {\paren {k + 1} x} + i \, \map \sin {\paren {k + 1} x} }$
=== Induction Step ===
This is our induction step:
Hence, by induction, for all $n \in \Z_{> 0}$:
:$\paren {r \paren {\cos x + i \sin x} }^n = r^n \paren {\map \cos {n x} + i \, \map \sin {n x} }$
== Sources ==
* : $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $20$
Category:De Moivre's Formula
Category:Proofs by Induction
De Moivre's Formula/Positive Integer Index/Proof 2

== Theorem ==
== Proof ==
From Product of Complex Numbers in Polar Form: General Result:
:$z_1 z_2 \cdots z_n = r_1 r_2 \cdots r_n \paren {\map \cos {\theta_1 + \theta_2 + \cdots + \theta_n} + i \, \map \sin {\theta_1 + \theta_2 + \cdots + \theta_n} }$
Setting $z_1 = z_2 = \cdots = z_n = r \paren {\cos x + i \sin x}$ gives the result.
== Sources ==
* : $\S 2$. Geometrical Representations: $(2.15)$
* : $1$: Complex Numbers: De Moivre's Theorem
Category:De Moivre's Formula
De Moivre's Formula/Proof 1

== Theorem ==
Let $z \in \C$ be a complex number expressed in complex form:
:$z = r \paren {\cos x + i \sin x}$
Then:
:$\forall \omega \in \C: \paren {r \paren {\cos x + i \sin x} }^\omega = r^\omega \map \cos {\omega x} + i \map \sin {\omega x}$
== Proof ==
Category:De Moivre's Formula
De Moivre's Formula/Rational Index

== Theorem ==
Let $z \in \C$ be a complex number expressed in complex form:
:$z = r \paren {\cos x + i \sin x}$
Then:
:$\forall p \in \Q: \paren {r \paren {\cos x + i \sin x} }^p = r^p \paren {\map \cos {p x} + i \, \map \sin {p x} }$
== Proof ==
Write $p = \dfrac a b$, where $a, b \in \Z$, $b \ne 0$.
Then:
== Also defined as ==
This result is also often presented in the simpler form:
:$\forall p \in \Q: \paren {\cos x + i \sin x}^p = \map \cos {p x} + i \, \map \sin {p x}$
== Also known as ==
De Moivre's Theorem.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: De Moivre's Theorem: $94 \ \text{(b)}$
Category:De Moivre's Formula
De Morgan's Laws (Boolean Algebras)

== Theorem ==
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra.
Then for all $a, b \in S$:
:$\neg \paren {a \vee b} = \neg a \wedge \neg b$
:$\neg \paren {a \wedge b} = \neg a \vee \neg b$
== Proof ==
By virtue of Complement in Boolean Algebra is Unique, it will suffice to verify:
:$\paren {a \vee b} \wedge \paren {\neg a \wedge \neg b} = \bot$
:$\paren {a \vee b} \vee \paren {\neg a \wedge \neg b} = \top$
For the first of these, compute:
By the Duality Principle, we also conclude:
:$\paren {a \wedge b} \vee \paren {\neg a \vee \neg b} = \top$
Substituting $\neg a$ and $\neg b$ for $a$ and $b$, respectively, this becomes:
:$\paren {\neg a \wedge \neg b} \vee \paren {a \vee b} = \top$
where Complement of Complement (Boolean Algebras) was applied.
Since $\vee$ is commutative, we have also derived the second of the two equations above.
Hence:
:$\neg \paren {a \vee b} = \neg a \wedge \neg b$
Now by again the Duality Principle:
:$\neg \paren {a \wedge b} = \neg a \vee \neg b$
follows immediately.
== Also see ==
* De Morgan's Laws (Logic)
* De Morgan's Laws (Predicate Logic)
* De Morgan's Laws (Set Theory)
== Historical Note ==
== Sources ==
* : $\S 2$: Exercise $2$
Category:Boolean Algebras
Category:De Morgan's Laws
De Morgan's Laws (Logic)

== Theorem ==
=== Disjunction of Negations ===
=== Conjunction of Negations ===
=== Conjunction ===
=== Disjunction ===
== The Intuitionist Perspective ==
Note that this:
can be proved in both directions without resorting to the LEM.
None of the others:
is reversible in intuitionistic logic.
== Also known as ==
== Also see ==
* De Morgan's Laws in the context of set theory.
== Historical Note ==
== Technical Note ==
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.4$: Statement Forms
* 
* 
Category:De Morgan's Laws (Logic)
Category:Conjunction
Category:Disjunction
De Morgan's Laws (Logic)/Conjunction/Definition

== Theorem ==
De Morgan's Law for Conjunction can be used as a definition for the logical operator that is conjunction:
:$p \land q := \neg \left({\neg p \lor \neg q}\right)$
== Historical Note ==
This definition appears in the introductory material of , and some other authors follow their lead.
== Sources ==
* : Chapter $\text{I}$: Preliminary Explanations of Ideas and Notations
Category:De Morgan's Laws (Logic)
De Morgan's Laws (Logic)/Conjunction/Formulation 1

== Theorem ==
:$p \land q \dashv \vdash \neg \paren {\neg p \lor \neg q}$
This can be expressed as two separate theorems:
=== Forward Implication ===
=== Reverse Implication ===
== Proof by Truth Table ==
== Sources ==
* : $\S 2.4$: Relations between Truth-Functions
* : Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Exercise $1 \ \text {(h)}$
* : $\S 2.3.3$
* : $\S 2.10$: Exercise $2.3$
Category:De Morgan's Laws (Logic)
De Morgan's Laws (Logic)/Conjunction/Formulation 1/Forward Implication

== Theorem ==
: $p \land q \vdash \neg \left({\neg p \lor \neg q}\right)$
== Proof ==
== Sources ==
* : $\S 1.2.5$: An aside: proof by contradiction: Exercises $1.6: \ 2 \ \text{(a)}$
Category:De Morgan's Laws (Logic)
De Morgan's Laws (Logic)/Conjunction/Formulation 1/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccc||cccccc|} \hline
p & \land & q & \neg & (\neg & p & \lor & \neg & q) \\
\hline
\F & \F & \F & \F & \T & \F & \T & \T & \F \\
\F & \F & \T & \F & \T & \F & \T & \F & \T \\
\T & \F & \F & \F & \F & \T & \T & \T & \F \\
\T & \T & \T & \T & \F & \T & \F & \F & \T \\
\hline
\end{array}$
Category:De Morgan's Laws (Logic)
Category:Truth Table Proofs
De Morgan's Laws (Logic)/Conjunction/Formulation 1/Reverse Implication

== Theorem ==
: $\neg \left({\neg p \lor \neg q}\right) \vdash p \land q$
== Proof ==
|}
== Sources ==
* : $\S 1.2.5$: An aside: proof by contradiction: Exercises $1.6: \ 2 \ \text{(b)}$
Category:De Morgan's Laws (Logic)
De Morgan's Laws (Logic)/Conjunction/Formulation 2

== Theorem ==
:$\vdash \paren {p \land q} \iff \paren {\neg \paren {\neg p \lor \neg q} }$
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : $\S 3.6$: Reference Formulae: $RF \, 14$
* : $\S 4.7$: The Derivation of Formulae: $D \, 24$
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T63}$
Category:De Morgan's Laws (Logic)
De Morgan's Laws (Logic)/Conjunction/Formulation 2/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations.
$\begin{array}{|ccc|c|cccccc|} \hline
(p & \land & q) & \iff & (\neg & (\neg & p & \lor & \neg & q)) \\
\hline
\F & \F & \F & \T & \F & \T & \F & \T & \T & \F \\
\F & \F & \T & \T & \F & \T & \F & \T & \F & \T \\
\T & \F & \F & \T & \F & \F & \T & \T & \T & \F \\
\T & \T & \T & \T & \T & \F & \T & \F & \F & \T \\
\hline
\end{array}$
== Sources ==
* : $\S \text{I}: 13$: Logical Equivalences: Exercise $\text{(A)} \ 3$
Category:De Morgan's Laws (Logic)
Category:Truth Table Proofs
De Morgan's Laws (Logic)/Conjunction of Negations

== Theorem ==
==== Formulation 1 ====
==== Formulation 2 ====
== Also see ==
* De Morgan's Laws (Set Theory) for a set theoretic application.
* De Morgan's Laws (Predicate Logic) for a predicate logic application.
== Sources ==
* 
* 
Category:De Morgan's Laws (Logic)
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1

== Theorem ==
:$\neg p \land \neg q \dashv \vdash \neg \paren {p \lor q}$
This can be expressed as two separate theorems:
=== Forward Implication ===
=== Reverse Implication ===
== Proof ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Exercise $1 \ \text {(f)}$
* : $\S 1.2.4$: Provable equivalence
Category:De Morgan's Laws (Logic)
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Forward Implication

== Theorem ==
:$\neg p \land \neg q \vdash \neg \paren {p \lor q}$
== Proof ==
== Sources ==
* : $\S 1.2.1$: Rules for natural deduction: Exercises $1.5: \ 2 \ \text{(a)}$
Category:De Morgan's Laws (Logic)
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccccc||cccc|} \hline
\neg & p & \land & \neg & q & \neg & (p & \lor & q) \\
\hline
\T & \F & \T & \T & \F & \T & \F & \F & \F \\
\T & \F & \F & \F & \T & \F & \F & \T & \T \\
\F & \T & \F & \T & \F & \F & \T & \T & \F \\
\F & \T & \F & \F & \T & \F & \T & \T & \T \\
\hline
\end{array}$
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Example $1.8 \ \text{(c)}$
Category:De Morgan's Laws (Logic)
Category:Truth Table Proofs