nyuuzyou/wikis huggingface dataset:
Modulus of Gamma Function of Imaginary Number/Also presented as

== Modulus of Gamma Function of Imaginary Number: Also presented as == 
Modulus of Gamma Function of Imaginary Number can also be presented as:
:$\cmod {\map \Gamma {i t} }^2 = \dfrac \pi {t \sinh \pi t}$
== Sources ==
* : $\S 16$: Miscellaneous Results: $16.17$
* : $\S 25$: The Gamma Function: Miscellaneous Results: $25.17.$
Category:Modulus of Gamma Function of Imaginary Number
Modulus of Gamma Function of One Half plus Imaginary Number

== Theorem == 
Let $t \in \R$ be a real number.
Then: 
:$\cmod {\map \Gamma {\dfrac 1 2 + i t} } = \sqrt {\pi \map \sech {\pi t} }$
where:
:$\Gamma$ is the Gamma function
:$\sech$ is the hyperbolic secant function.
== Proof == 
As $\cmod z \ge 0$ for all complex numbers $z$, we can take the non-negative square root of both sides and write:
:$\cmod {\map \Gamma {\dfrac 1 2 + i t} } = \sqrt {\pi \map \sech {\pi t} }$
Category:Gamma Function
Category:Hyperbolic Secant Function
Modulus of Limit

== Theorem ==
Let $X$ be one of the standard number fields $\Q, \R, \C$.
Let $\sequence {x_n}$ be a sequence in $X$.
Let $\sequence {x_n}$ be convergent to the limit $l$.
That is, let $\ds \lim_{n \mathop \to \infty} x_n = l$.
Then
:$\ds \lim_{n \mathop \to \infty} \cmod {x_n} = \cmod l$
where $\cmod {x_n}$ is the modulus of $x_n$.
=== Normed Division Ring ===
=== Normed Vector Space ===
== Proof ==
By the Triangle Inequality, we have:
:$\cmod {\cmod {x_n} - \cmod l} \le \cmod {x_n - l}$
Hence by the Squeeze Theorem and Convergent Sequence Minus Limit, $\cmod {x_n} \to \cmod l$ as $n \to \infty$.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (1) \ \text{(ii)}$
Category:Limits of Sequences
Category:Modulus of Limit
Modulus of Limit/Normed Division Ring

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$ be a convergent sequence in $R$ to the limit $l$.
That is, let $\ds \lim_{n \mathop \to \infty} x_n = l$.
Then
:$\ds \lim_{n \mathop \to \infty} \norm {x_n} = \norm l$
where $\sequence {\norm {x_n} }$ is a real sequence.
== Proof ==
By the Reverse Triangle Inequality, we have:
:$\cmod {\norm {x_n} - \norm l} \le \norm {x_n - l}$
Hence by the Squeeze Theorem and Convergent Sequence Minus Limit, $\norm {x_n} \to \norm l$ as $n \to \infty$.
== Sources ==
Category:Modulus of Limit
Category:Normed Division Rings
Modulus of Limit/Normed Vector Space

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\sequence {x_n}$ be a convergent sequence in $R$ to the limit $x$.
That is, let $\ds \lim_{n \mathop \to \infty} x_n = x$.
Then:
:$\ds \lim_{n \mathop \to \infty} \norm {x_n} = \norm x$
where $\sequence {\norm {x_n} }$ is a real sequence.
== Proof ==
By the Reverse Triangle Inequality:
:$\cmod {\norm {x_n} - \norm x} \le \norm {x_n - x}$
Hence by the Squeeze Theorem and Convergent Sequence Minus Limit:
:$\norm {x_n} \to \norm x$
as $n \to \infty$.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Modulus of Limit
Category:Convergent Sequences (Normed Vector Spaces)
Modulus of Linear Functional on Vector Space is Seminorm

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$.
Let $f : X \to \GF$ be a linear functional.
Define $p_f : X \to \R_{\ge 0}$ by: 
:$\map {p_f} x = \cmod {\map f x}$ 
for each $x \in X$.
Then $p_f$ is a seminorm. 
== Proof ==
=== Proof of ===
For each $\lambda \in \GF$ and $x \in X$, we have: 
=== Proof of ===
For each $x, y \in X$, we have:
Category:Seminorms
Modulus of Positive Real Number to Complex Power is Positive Real Number to Power of Real Part

== Theorem ==
Let $z \in \C$ be a complex number.
Let $t > 0$ be wholly real.
Let $t^z$ be $t$ to the power of $z$ defined on its principal branch.
Then:
:$\cmod {t^z} = t^{\map \Re z}$
== Proof ==
 
Category:Complex Modulus
Category:Complex Analysis
Modulus of Sine of Complex Number

== Theorem ==
Let $z = x + i y \in \C$ be a complex number, where $x, y \in \R$.
Let $\sin z$ denote the complex sine function.
Then:
:$\cmod {\sin z} = \sqrt {\sin^2 x + \sinh^2 y}$
where:
:$\cmod z$ denotes the modulus of a complex number $z$
:$\sin x$ denotes the real sine function
:$\sinh$ denotes the hyperbolic sine function.
== Proof ==
== Sources ==
* : $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$
Category:Sine Function
Modulus of Sine of x Less Than or Equal To Absolute Value of x

== Theorem ==
Let $x$ be a real number.
Then:
:$\size {\sin x} \le \size x$
== Proof ==
Clearly the inequality holds if $x = 0$.
Take $x \ne 0$. 
From the Mean Value Theorem and Derivative of Sine Function, there exists $c \in \R$ such that: 
:$\ds \frac {\sin x - \sin 0} {x - 0} = \cos c$
so:
:$\sin x = x \cos c$
Then we have: 
as required.
Category:Inequalities
Category:Sine Function
Category:Absolute Value Function
Modulus of Sum equals Modulus of Distance implies Quotient is Imaginary

== Theorem ==
Let $z_1$ and $z_2$ be complex numbers such that:
:$\cmod {z_1 + z_2} = \cmod {z_1 - z_2}$
Then $\dfrac {z_2} {z_1}$ is wholly imaginary.
=== Corollary ===
== Proof ==
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$.
Then:
Now we have:
But we have:
:$x_1 x_2 + y_1 y_2 = 0$
Thus:
:$\dfrac {z_1} {z_2} = \dfrac {i \paren {x_2 y_1 - x_1 y_2} } { {x_2}^2 + {y_2}^2}$
which is wholly imaginary.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $147 \ \text{(a)}$
Category:Complex Modulus
Modulus of Sum equals Modulus of Distance implies Quotient is Imaginary/Corollary

== Corollary to Modulus of Sum equals Modulus of Distance implies Quotient is Imaginary ==
Let $P_1$ and $P_2$ be points embedded in the complex plane.
Let $P_1$ and $P_2$ be represented by the complex numbers $z_1$ and $z_2$ be complex numbers such that:
:$\cmod {z_1 + z_2} = \cmod {z_1 - z_2}$
Then:
:$\angle P_1 O P_2 = 90 \degrees$
== Proof ==
Let $z_1 = x_1 + i y_1 = r_1 e^{i \theta_1}, z_2 = x_2 + i y_2 = r_2 e^{i \theta_2}$.
We have from Division of Complex Numbers in Polar Form that:
:$\dfrac {z_1} {z_2} = \dfrac {r_1} {r_2} \paren {\map \cos {\theta_1 - \theta_2} + i \, \map \sin {\theta_1 - \theta_2} }$
From Modulus of Sum equals Modulus of Distance implies Quotient is Imaginary, $\dfrac {z_1} {z_2}$ is wholly imaginary.
Hence $\map \cos {\theta_1 - \theta_2} = 0$.
Hence from Cosine of Half-Integer Multiple of Pi, $\theta_1 - \theta_2 = \paren {n + \dfrac 1 2} \pi$ for some $n$.
In the geometrical context, that means:
:$\theta_1 - \theta_2 = 90 \degrees$
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $147 \ \text{(b)}$
Category:Complex Modulus
Modulus z - 1 Less than Modulus z + 1 iff Real z Greater than Zero

== Theorem ==
Let $z \in \C$ be a complex number.
Then:
:$\cmod {z - 1} < \cmod {z + 1} \iff \map \Re z > 0$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $144$
Category:Complex Modulus
Modus Ponendo Ponens

== Proof Rule ==
Modus ponendo ponens is a valid argument in types of logic dealing with conditionals $\implies$.
This includes propositional logic and predicate logic, and in particular natural deduction.
=== Proof Rule ===
=== Sequent Form ===
== Variants ==
The following forms can be used as variants of this theorem:
=== Variant 1 ===
=== Variant 2 ===
=== Variant 3 ===
== Examples ==
== Also known as ==
== Also see ==
* Definition:Conditional
* Rule of Implication
The following are related argument forms:
* Modus Ponendo Tollens
* Modus Tollendo Ponens
* Modus Tollendo Tollens
== Linguistic Note ==
== Sources ==
* : Chapter $\text{I}$: Preliminary Explanations of Ideas and Notations
* : $2$ Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic
* : $\S 1.12$: Valid Arguments
* 
* 
* 
* 
* : $\text{II}.10$ Formal Proofs
Category:Modus Ponendo Ponens
Category:Conditional
Modus Ponendo Ponens/Also known as

== Proof Rule ==
Modus Ponendo Ponens is also known as:
* Modus Ponens, abbreviated M.P.
* The Rule of Implies-Elimination
* The Rule of Arrow-Elimination
* The Rule of Material Detachment, or just Rule of Detachment
* The Process of Inference
, seemingly uneasy with the language they are using, state:
:The process of the inference cannot be reduced to symbols.
Having said that, they then go on to write:
:... we shall write instead
::"$\vdash p \supset \, \vdash q$,"
:which is to be considered as a mere abbreviation of the threefold statement
::"$\vdash p$" and "$\vdash \paren {p \supset q}$" and "$\vdash q$."
Remember that use $\supset$ to denote the implication function.
== Sources ==
* : Chapter $\text{I}$: Preliminary Explanations of Ideas and Notations
* : $3$: The Method of Deduction: $3.1$: Formal Proof of Validity: Rules of Inference: $1.$
* 
* 
* 
* 
Category:Modus Ponendo Ponens
Modus Ponendo Ponens/Examples/Jones is Mortal

== Examples of Modus Ponendo Ponens ==
The following is an example of use of Modus Ponendo Ponens:
:If Jones is a man, then Jones is mortal.
:Jones is a man.
:Therefore, Jones is Mortal.
== Sources ==
* 
* 
Category:Examples of Modus Ponendo Ponens
Modus Ponendo Ponens/Proof Rule

== Proof Rule ==
Modus ponendo ponens is a valid argument in types of logic dealing with conditionals $\implies$.
This includes propositional logic and predicate logic, and in particular natural deduction.
As a proof rule it is expressed in the form:
:If we can conclude $\phi \implies \psi$, and we can also conclude $\phi$, then we may infer $\psi$.
Thus it provides a means of eliminating a conditional from a sequent.
It can be written:
:$\ds {\phi \qquad \phi \implies \psi \over \psi} \to_e$
=== Tableau Form ===
== Also known as ==
== Also see ==
* This is a rule of inference of the following proof systems:
** Definition:Natural Deduction
** Definition:Hilbert Proof System Instance 1 for Predicate Logic
** Definition:Hilbert Proof System/Instance 1
** Definition:Hilbert Proof System/Instance 2
== Linguistic Note ==
== Technical Note ==
== Sources ==
* : $\S \text{II}.15$: Rules of inference
* : $\S 3.9$: Derivation by Substitution
* : $\S 4.2$: The Construction of an Axiom System: $RST \, 3$
* : $\text{I}$: 'NOT' and 'IF': $\S 3$
* : Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation
* : $\S 1$: Some mathematical language: Axioms
* : $\S \text{II}$: The Logic of Statements $(2): \ 1$: Decision procedures and proofs: $1$
* : $\S 1.2.1$: Rules for natural deduction
Category:Rules of Inference
Category:Modus Ponendo Ponens
Modus Ponendo Ponens/Proof Rule/Tableau Form

== Proof Rule ==
Let $\phi \implies \psi$ be a well-formed formula in a tableau proof whose main connective is the implication operator.
The Modus Ponendo Ponens is invoked for $\phi \implies \psi$ and $\phi$ as follows:
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation
Category:Tableau Proofs
Category:Modus Ponendo Ponens
Modus Ponendo Ponens/Proof Rule/Technical Note

== Technical Note on ==
When invoking in a tableau proof, use the template:
:{{ModusPonens|line|pool|statement|first|second}}
or:
:{{ModusPonens|line|pool|statement|first|second|comment}}
where:
:line is the number of the line on the tableau proof where is to be invoked
:pool is the pool of assumptions (comma-separated list)
:statement is the statement of logic that is to be displayed in the Formula column, without the $ ... $ delimiters
:first is the first of the two lines of the tableau proof upon which this line directly depends, the one in the form $p \implies q$
:second is the second of the two lines of the tableau proof upon which this line directly depends, the one in the form $p$
:comment is the (optional) comment that is to be displayed in the Notes column.
Category:Technical Notes
Modus Ponendo Ponens/Sequent Form

== Theorem ==
The Modus Ponendo Ponens can be symbolised by the sequent:
== Proof 1 ==
== Proof 2 ==
== Proof by Truth Table ==
== Sources ==
* : $3$: The Method of Deduction: $3.1$: Formal Proof of Validity: Rules of Inference: $1.$
Category:Modus Ponendo Ponens
Modus Ponendo Ponens/Sequent Form/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
$\begin{array}{|c|ccc||c|} \hline
p & p & \implies & q & q\\
\hline
\F & \F & \T & \F & \F \\
\F & \F & \T & \T & \T \\
\T & \T & \F & \F & \F \\
\T & \T & \T & \T & \T \\
\hline
\end{array}$
As can be seen, when $p$ is true, and so is $p \implies q$, then $q$ is also true.
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables
* : $\S 1$: Informal statement calculus: $\S 1.3$: Rules for manipulation and substitution: Proposition $1.9$
Category:Truth Table Proofs
Category:Modus Ponendo Ponens
Modus Ponendo Ponens/Variant 1

== Theorem ==
:$p \vdash \paren {p \implies q} \implies q$
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation: Exercise $1 \ \text{(m)}$
Category:Modus Ponendo Ponens
Modus Ponendo Ponens/Variant 1/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
$\begin{array}{|c|ccccc|} \hline
p & (p & \implies & q) & \implies & q\\
\hline
\F & \F & \T & \F & \F & \F \\
\F & \F & \T & \T & \T & \T \\
\T & \T & \F & \F & \T & \F \\
\T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
As can be seen by inspection, when $p$ is true, the value of the main connective is also true.
Category:Truth Table Proofs
Category:Modus Ponendo Ponens
Modus Ponendo Ponens/Variant 2/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
$\begin{array}{|c|c|ccccc|} \hline
p & \implies & ((p & \implies & q) & \implies & q)\\
\hline
\F & \T & \F & \T & \F & \F & \F \\
\F & \T & \F & \T & \T & \T & \T \\
\T & \T & \T & \F & \F & \T & \F \\
\T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
As can be seen by inspection, the main connective is true throughout.
Category:Truth Table Proofs
Category:Modus Ponendo Ponens
Modus Ponendo Ponens/Variant 3

== Theorem ==
:$\vdash \paren {\paren {p \implies q} \land p} \implies q$
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : $\S 3.6$: Reference Formulae: $RF \, 20$
Category:Modus Ponendo Ponens
Modus Ponendo Ponens/Variant 3/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.
$\begin{array}{|ccccc|c|c|} \hline
((p & \implies & q) & \land & p) & \implies & q \\
\hline
\F & \T & \F & \F & \F & \T & \F \\
\F & \T & \T & \F & \F & \T & \T \\
\T & \F & \F & \F & \T & \T & \F \\
\T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
== Sources ==
* : $\S \text{I}: 10$: Testing the Validity of Arguments
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic
Category:Truth Table Proofs
Category:Modus Ponendo Ponens
Modus Ponendo Ponens for Semantic Consequence in Predicate Logic

== Theorem ==
Let $\mathrm{PL}$ be the formal semantics of structures for predicate logic.
Denote with $\models_{\mathrm{PL}}$ $\mathrm{PL}$-semantic consequence.
Let $\mathbf A$ and $\mathbf B$ be sentences.
Let $\FF$ be a set of sentences.
Suppose that:
:$\FF \models_{\mathrm{PL}} \mathbf A$
:$\FF \models_{\mathrm{PL}} \mathbf A \implies \mathbf B$
Then:
:$\FF \models_{\mathrm{PL}} \mathbf B$
establishing Modus Ponendo Ponens in $\mathrm{PL}$.
== Proof ==
To establish $\FF \models_{\mathrm{PL}} \mathbf B$, we need to show that for all structures $\AA$:
:$\AA \models_{\mathrm{PL}} \FF$ implies $\AA \models_{\mathrm{PL}} \mathbf B$
where $\models_{\mathrm{PL}}$ denotes the models relation.
Then since $\AA \models_{\mathrm{PL}} \FF$, it follows that:
:$\AA \models_{\mathrm{PL}} \mathbf A$
:$\AA \models_{\mathrm{PL}} \mathbf A \implies \mathbf B$
By definition then, the latter means:
:$\map {f^\to} { \map {\operatorname{val}_\AA} {\mathbf A}, \map {\operatorname{val}_\AA} {\mathbf B} } = T$
where $f^\to$ is the truth function of $\implies$ and $\map {\operatorname{val}_\AA} {\mathbf A}$ is the value of $\mathbf A$ in $\AA$.
Since $\map {\operatorname{val}_\AA} {\mathbf A} = T$, it then follows from the definition of $f^\to$ that necessarily:
:$\map {\operatorname{val}_\AA} {\mathbf B} = T$
which is to say:
:$\AA \models_{\mathrm{PL}} \mathbf B$
Since $\AA$ was arbitrary such that $\AA \models_{\mathrm{PL}} \FF$, it follows that:
:$\FF \models_{\mathrm{PL}} \mathbf B$
Category:Model Theory for Predicate Logic
Modus Ponendo Tollens

== Modus Ponendo Tollens ==
Modus ponendo tollens is a valid argument in types of logic dealing with conjunctions $\land$ and negation $\neg$.
This includes propositional logic and predicate logic, and in particular natural deduction.
=== Proof Rule ===
== Variants ==
The following forms can be used as variants of this theorem:
=== Variant ===
== Explanation ==
== Linguistic Note ==
== Also see ==
* Definition:Conjunction
The following are related argument forms:
* Modus Ponendo Ponens
* Modus Tollendo Ponens
* Modus Tollendo Tollens
Category:Conjunction
Category:Logical Negation
Category:Modus Ponendo Tollens
Modus Ponendo Tollens/Proof Rule

== Proof Rule ==
Modus ponendo tollens is a valid argument in types of logic dealing with conjunctions $\land$ and negation $\neg$.
This includes propositional logic and predicate logic, and in particular natural deduction.
As a proof rule it is expressed in either of the two forms:
:$(1): \quad$ If we can conclude $\map \neg {\phi \land \psi}$, and we can also conclude $\phi$, then we may infer $\neg \psi$.
:$(2): \quad$ If we can conclude $\map \neg {\phi \land \psi}$, and we can also conclude $\psi$, then we may infer $\neg \phi$.
It can be written:
:$\ds {\map \neg {\phi \land \psi} \quad \phi \over \neg \psi} \textrm{MPT}_1 \qquad \text{or} \qquad {\map \neg {\phi \land \psi} \quad \psi \over \neg \phi} \textrm{MPT}_2$
=== Tableau Form ===
=== Explanation ===
== Also see ==
* This is a rule of inference of the following proof systems:
** Definition:Natural Deduction
== Linguistic Note ==
== Technical Note ==
Category:Rules of Inference
Category:Modus Ponendo Tollens
Modus Ponendo Tollens/Proof Rule/Tableau Form

== Tableau Form of Modus Ponendo Tollens ==
Let $\phi \land \psi$ be a well-formed formula in a tableau proof whose main connective is the conjunction operator.
The Modus Ponendo Tollens is invoked for $\neg \left({\phi \land \psi}\right)$ in either of the two forms:
;Form 1:
;Form 2:
Category:Modus Ponendo Tollens
Category:Tableau Proofs
Modus Ponendo Tollens/Proof Rule/Technical Note

== Technical Note on ==
When invoking in a tableau proof, use the template:
:{{ModusPonendoTollens|line|pool|statement|first|second|1 or 2}}
or:
:{{ModusPonendoTollens|line|pool|statement|first|second|1 or 2|comment}}
where:
:line is the number of the line on the tableau proof where is to be invoked
:pool is the pool of assumptions (comma-separated list)
:statement is the statement of logic that is to be displayed in the Formula column, without the $ ... $ delimiters
:first is the first of the two lines of the tableau proof upon which this line directly depends, the one in the form $\neg \left({\phi \land \psi}\right)$
:second is the second of the two lines of the tableau proof upon which this line directly depends, the one in the form $\phi$ or $\psi$
:1 or 2 should hold 1 for ModusPonendoTollens_1, and 2 for ModusPonendoTollens_2
:comment is the (optional) comment that is to be displayed in the Notes column.
Category:Technical Notes
Modus Ponendo Tollens/Sequent Form/Case 1

== Theorem ==
:$\neg \paren {p \land q}, p \vdash \neg q$
== Proof ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Ruless: Theorem $34$
Category:Modus Ponendo Tollens
Modus Ponendo Tollens/Variant/Formulation 1

== Theorem ==
If two statements can not both be true, and one of them is true, it follows that the other one is not true.
: $\neg \left({p \land q}\right) \dashv \vdash p \implies \neg q$
This theorem is known as the modus ponendo tollens.
Its abbreviation in a tableau proof is $\mathrm {MPT}$.
This can be expressed as two separate theorems:
=== Forward Implication ===
=== Reverse Implication ===
== Proof ==
Category:Modus Ponendo Tollens
Modus Ponendo Tollens/Variant/Formulation 1/Proof

== Theorem ==
:$\neg \paren {p \land q} \dashv \vdash p \implies \neg q$
== Proof ==
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|cccc||cccc|} \hline
\neg & (p & \land & q) & p & \implies & \neg & q \\
\hline
T & F & F & F & F & T & T & F \\
T & F & F & T & F & T & F & T \\
T & T & F & F & T & T & T & F \\
F & T & T & T & T & F & F & T \\
\hline
\end{array}$
Category:Modus Ponendo Tollens
Category:Truth Table Proofs
Modus Ponendo Tollens/Variant/Formulation 2

== Theorem ==
:$\vdash \paren {\neg \paren {p \land q} } \iff \paren {p \implies \neg q}$
== Proof ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$: Theorem $\text{T39}$
Category:Modus Ponendo Tollens
Modus Tollendo Ponens

== Sequent ==
Modus tollendo ponens is a valid argument in types of logic dealing with disjunctions $\lor$ and negation $\neg$.
This includes propositional logic and predicate logic, and in particular natural deduction.
=== Proof Rule ===
=== Sequent Form ===
== Variants ==
The following forms can be used as variants of this theorem:
=== Variant ===
Note that the form:
requires Law of Excluded Middle.
Therefore it is not valid in intuitionistic logic.
== Explanation ==
== Also known as ==
== Examples ==
== Also see ==
* Definition:Disjunction
The following are related argument forms:
* Modus Ponendo Ponens
* Modus Ponendo Tollens
* Modus Tollendo Tollens
== Linguistic Note ==
== Sources ==
* 
* 
* 
Category:Modus Tollendo Ponens
Category:Disjunction
Category:Logical Negation
Modus Tollendo Ponens/Also known as

== Sequent ==
The Modus Tollendo Ponens is also known as the Disjunctive Syllogism, abbreviated D.S.
== Sources ==
* : $3$: The Method of Deduction: $3.1$: Formal Proof of Validity: Rules of Inference: $4.$
* 
* 
* 
Category:Modus Tollendo Ponens
Modus Tollendo Ponens/Examples/Third World War

== Examples of Use of Modus Tollendo Ponens ==
The following is an example of the use of a Modus Tollendo Ponens:
:The United Nations will be strengthened or there will be a third world war.
:The United Nations will not be strengthened.
::Therefore there will be a third world war.
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.1$: Simple and Compound Statements
Category:Modus Tollendo Ponens
Modus Tollendo Ponens/Explanation

== Proof Rule ==
The Modus Tollendo Ponens can be expressed in natural language as:
If either of two statements is true, and one of them is not to be true, it follows that the other one is true.
:Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth.
:::-- Sherlock Holmes (Sir Arthur Conan Doyle, 1890: The Sign of the Four, ch. 6)
Category:Modus Tollendo Ponens
Modus Tollendo Ponens/Mistake

== Source Work ==
:
:modus tollendo ponens
== Mistake ==
:[Latin: method of denying and affirming]
== Correction ==
The above is an oversimplification that leads to an inaccuracy.
Technically speaking it actually means method of affirming by denying.
== Sources ==
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Category:Mistakes/The Penguin Dictionary of Mathematics