nyuuzyou/wikis huggingface dataset:
Multiplicatively Closed Subset is Saturated iff Complement is Union of Prime Ideals

== Definition ==
Let $A$ be a commutative ring with unity.
Let $S \subseteq A$ be a multiplicatively closed subset.
Then the following are equivalent:
:$(1):\quad$ $S$ is saturated
:$(2):\quad$ $\relcomp A S = \bigcup \set {\mathfrak p \in \Spec A : \mathfrak p \cap S = \O }$
:$(3):\quad$ $\exists \TT \subseteq \Spec A : \relcomp A S = \bigcup \TT$
where:
:$\relcomp A S$ denotes the complement of $S$
:$\Spec A$ denotes the prime spectrum of $A$
== Proof ==
=== $(2) \implies (3)$ ===
This is clear, choosing:
:$\TT = \set {\mathfrak p \in \Spec A : \mathfrak p \cap S = \O }$.
=== $(3) \implies (1)$ ===
Let $\TT \subseteq \Spec A$ be such that:
:$\relcomp A S = \bigcup \TT$.
Let $x, y \in A$ be such that $x y \in S$.
That is:
:$\forall \mathfrak p \in \TT : xy \not \in \mathfrak p$
As $\mathfrak p$s are ideals, we have:
:$\forall \mathfrak p \in \TT : x \not \in \mathfrak p \land y \not \in \mathfrak p$
That is:
:$x, y \in S$
=== $(1) \implies (2)$ ===
Category:Localization of Rings
Multiplicity of 720 in 720 Factorial

== Theorem ==
The multiplicity of $720$ in $720!$ is $178$.
That is:
:$720^{178} \divides 720!$
but:
:$720^{179} \nmid 720!$
where:
:$720!$ denotes $720$ factorial
:$\divides$ denotes divisibility
:$\nmid$ denotes non-divisibility.
== Proof ==
We have that:
:$720 = 2^4 \times 3^2 \times 5$
It remains to inspect the divisibility of $2$, $3$ and $5$ in $720!$
Thus:
=== Multiplicity of $2$ in $720!$ ===
=== Multiplicity of $3$ in $720!$ ===
=== Multiplicity of $5$ in $720!$ ===
We calculate the multiplicity of the powers of $2$ and $3$ in $720!$ thus:
Thus it is seen that the smallest power of the prime powers that are divisors of $720$ that divide $720!$ is that of $3^2$, which is $178$.
Hence:
:$720^{178} \divides 720!$
but:
:$720^{179} \nmid 720!$
== Sources ==
* : $720$
Category:Factorials
Category:720
Category:De Polignac's Formula
Multiply Perfect Number of Order 6

== Theorem ==
The number defined as:
:$n = 2^{36} \times 3^8 \times 5^5 \times 7^7 \times 11 \times 13^2 \times 19 \times 31^2$
::$\times \ 43 \times 61 \times 83 \times 223 \times 331 \times 379 \times 601 \times 757 \times 1201$
::$\times \ 7019 \times 112 \, 303 \times 898 \, 423 \times 616 \, 318 \, 177$
is multiply perfect of order $6$.
== Proof ==
From Divisor Sum Function is Multiplicative, we may take each prime factor separately and form $\map {\sigma_1} n$ as the product of the divisor sum of each.
Each of the prime factors which occur with multiplicity $1$ will be treated first.
A prime factor $p$ contributes towards the combined $\sigma_1$ a factor $p + 1$.
Hence we have:
The remaining factors are treated using Divisor Sum of Power of Prime:
:$\map {\sigma_1} {p^k} = \dfrac {p^{k + 1} - 1} {p - 1}$
Thus:
Gathering up the prime factors, we have:
:$\map {\sigma_1} n = 2^{37} \times 3^9 \times 5^5 \times 7^7 \times 11 \times 13^2 \times 19 \times 31^2$
::$\times \ 43 \times 61 \times 83 \times 223 \times 331 \times 379 \times 601 \times 757 \times 1201$
::$\times \ 7019 \times 112 \, 303 \times 898 \, 423 \times 616 \, 318 \, 177$
By inspection of the multiplicities of the prime factors of $n$ and $\map {\sigma_1} n$, it can be seen that they match for all except for $2$ and $3$.
It follows that $\map {\sigma_1} n = 2 \times 3 \times n = 6 n$.
Hence the result.
== Historical Note ==
== Sources ==
* : $100,895,598,169$
* : $100,895,598,169$
Category:Multiply Perfect Numbers
Multiply Perfect Number of Order 6/Historical Note

== Historical Note on Multiply Perfect Number of Order 6 ==
, in a letter of $1643$, challenged to find the ratio of:
:$2^{36} \times 3^8 \times 5^5 \times 11 \times 13^2 \times 19 \times 31^2$
::$\times \ 43 \times 61 \times 83 \times 223 \times 331 \times 379 \times 601 \times 757 \times 1201$
::$\times \ 7019 \times 823 \, 543 \times 616 \, 318 \, 177 \times 100 \, 895 \, 598 \, 169$
to its aliquot sum.
 replied that its ratio to the sum of all its divisors (including the number itself) was $6$.
He also pointed out that $100 \, 895 \, 598 \, 169 = 112 \, 303 \times 898 \, 423$, both of which divisors are prime.
Also note that $823 \, 543 = 7^7$, another point that glossed over, intentionally or inadvertently, in his initial challenge.
Both 's initial challenge and 's factorisation of $100 \, 895 \, 598 \, 169$ were remarkable, considering the lack of computing machines in those days. To this day, nobody knows how they did it.
== Sources ==
* : $100,895,598,169$
* : Chapter $\text {A}.13$: Fermat ()
* : $100,895,598,169$
Category:Historical Notes
Multiply Perfect Number of Order 8

== Theorem ==
The number defined as:
:$n = 2^{65} \times 3^{23} \times 5^9 \times 7^{12} \times 11^3 \times 13^3 \times 17^2 \times 19^2 \times 23 \times 29^2 \times 31^2$
::$\times 37 \times 41 \times 53 \times 61 \times 67^2 \times 71^2 \times 73 \times 83 \times 89 \times 103 \times 127 \times 131$
::$\times 149 \times 211 \times 307 \times 331 \times 463 \times 521 \times 683 \times 709 \times 1279 \times 2141 \times 2557 \times 5113$
::$\times 6481 \times 10 \, 429 \times 20 \, 857 \times 110 \, 563 \times 599 \, 479 \times 16 \, 148 \, 168 \, 401$
is multiply perfect of order $8$.
== Proof ==
From Divisor Sum Function is Multiplicative, we may take each prime factor separately and form $\map {\sigma_1} n$ as the product of the divisor sum of each.
Each of the prime factors which occur with multiplicity $1$ will be treated first.
A prime factor $p$ contributes towards the combined $\sigma_1$ a factor $p + 1$.
Hence we have:
The remaining factors are treated using Divisor Sum of Power of Prime:
:$\map {\sigma_1} {p^k} = \dfrac {p^{k + 1} - 1} {p - 1}$
Thus:
Gathering up the prime factors, we have:
:$\map {\sigma_1} n = 2^{68} \times 3^{23} \times 5^9 \times 7^{12} \times 11^3 \times 13^3 \times 17^2 \times 19^2 \times 23 \times 29^2 \times 31^2$
::$\times 37 \times 41 \times 53 \times 61 \times 67^2 \times 71^2 \times 73 \times 83 \times 89 \times 103 \times 127 \times 131$
::$\times 149 \times 211 \times 307 \times 331 \times 463 \times 521 \times 683 \times 709 \times 1279 \times 2141 \times 2557 \times 5113$
::$\times 6481 \times 10 \, 429 \times 20 \, 857 \times 110 \, 563 \times 599 \, 479 \times 16 \, 148 \, 168 \, 401$
By inspection of the multiplicities of the prime factors of $n$ and $\map {\sigma_1} n$, it can be seen that they match for all except for $2$.
It follows that $\map {\sigma_1} n = 2^3 \times n = 8 n$.
Hence the result.
== Sources ==
* 
* : $120$
* 
* : $120$
Category:Multiply Perfect Numbers
Multiply Perfect Number of Order 8/Mistake

== Source Work ==
:
:The Dictionary
::$120$
:
:The Dictionary
::$120$
== Mistake ==
:One of the smallest [multiply perfect numbers] of order $8$ was discovered by , an American 'human computer': $2 \times 3^{23} \times 5^9 \times 7^{12} \times 11^3 \times 13^3 \times 17^2 \times$
:$19^2 \times 23 \times 29^2 \times 31^2 \times 37 \times 41 \times 53 \times 61 \times 67^2 \times 71^2 \times 73 \times 83$
:$\times 89 \times 103 \times 127 \times 131 \times 149 \times 211 \times 307 \times 331 \times 463 \times 521$
:$\times 683 \times 709 \times 1279 \times 2141 \times 2557 \times 5113 \times 6481 \times 10,429$
:$\times 20,857 \times 110,563 \times 599,479 \times 16,148,168,401$.
The multiplicity of $2$ has been omitted: the expression should read $2^{65} \times 3^{23} \times 5^9 \ldots$
In ' defence, he merely propagates the error as it appears in 's of $1994$.
== Sources ==
* : $120$
* 
* : $120$
Category:Mistakes/Curious and Interesting Numbers
Mutual Order Embedding does not imply Order Isomorphism

== Theorem ==
Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be ordered sets.
Let it be possible for:
:$\struct {S_1, \preceq_1}$ to be embedded in $\struct {S_2, \preceq_2}$
:$\struct {S_2, \preceq_2}$ to be embedded in $\struct {S_1, \preceq_1}$.
Then it is not necessarily the case that $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ are isomorphic.
== Proof ==
Consider the ordered structures:
:$\struct {S_1, \preceq_1} := \struct {\R, \le}$
:$\struct {S_2, \preceq_2} := \struct {\hointl {-\dfrac \pi 2} {\dfrac \pi 2}, \le}$
From Real Arctangent Function is Order Embedding into Reals, $\struct {S_1, \preceq_1}$ can be embedded into $\struct {S_2, \preceq_2}$.
From Inclusion Mapping is Order Embedding, $\struct {S_1, \preceq_2}$ can be embedded into $\struct {S_2, \preceq_1}$.
 $\struct {\R, \le}$ and $\struct {\hointl {-\dfrac \pi 2} {\dfrac \pi 2}, \le}$ are isomorphic.
From Number of Maximal Elements is Order Property, $\struct {\R, \le}$ and $\struct {\hointl {-\dfrac \pi 2} {\dfrac \pi 2}, \le}$ have the same number of maximal elements.
But we note that:
:$\struct {\R, \le}$ has no maximal elements
:$\struct {\hointl {-\dfrac \pi 2} {\dfrac \pi 2}, \le}$ has one maximal element, that is $\dfrac \pi 2$.
It follows by Proof by Contradiction that $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ are not isomorphic.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $22$
Category:Order Embeddings
Category:Order Isomorphisms
Möbius Function/Examples/1

== Example of Möbius Function ==
For $n \in \Z_{>0}$, let $\map \mu n$ denote the Möbius function of $n$.
Then:
:$\map \mu 1 = 1$
== Proof ==
By definition of the Möbius function:
:$\map \mu 1 = 1$
== Sources ==
* 
Category:Examples of Möbius Function
Möbius Function is Multiplicative

== Theorem ==
The Möbius function $\mu$ is a multiplicative function:
:$m \perp n \implies \map \mu {m n} = \map \mu m \map \mu n$
where $m, n \in \Z_{>0}$.
=== Corollary ===
== Proof ==
First note that we have $\map \mu 1 = 1$, which agrees with Value of Multiplicative Function at One.
Let $m, n \in \Z_{>0}$ such that $m \perp n$.
First, suppose that either $\map \mu m = 0$ or $\map \mu n = 0$.
Then either $m$ or $n$ has a factor $p^2$ where $p$ is prime.
Thus it will follow that $m n$ will also have a factor $p^2$ and hence $\map \mu {m n} = 0$.
So the result holds when $\map \mu m = 0$ or $\map \mu n = 0$.
Now suppose that $\map \mu m \ne 0$ and $\map \mu n \ne 0$.
Let $m = p_1 p_2 \ldots p_r$, $n = q_1 q_2 \ldots q_s$ where all the $p_i, q_j$ are prime.
Then:
:$m n = p_1 p_2 \ldots p_r q_1 q_2 \ldots q_s$
As $m \perp n$ it follows that:
:$\forall i, j: p_i \ne q_j$
Hence there is no prime in $m n$ whose power is higher than $1$, which means that $\map \mu {m n} \ne 0$.
So:
:$\map \mu {m n} = \paren {-1}^{r + s} = \paren {-1}^r \paren {-1}^s = \map \mu m \map \mu n$
Hence the result.
== Sources ==
* : Chapter $1$: Properties of the Natural Numbers: $\S 25 \beta$
* 
Category:Möbius Function
Category:Multiplicative Functions
Möbius Function is Multiplicative/Corollary

== Theorem ==
Let $\gcd \set {m, n} > 1$.
Then:
:$\map \mu {m n} = 0$
where $\mu$ denotes the Möbius function.
== Proof ==
Let $\gcd \set {m, n} = k$ where $k > 1$.
Then $m = k r$ and $n = k s$ for some $r, s \in \Z$.
Thus $m n = k^2 r s$.
From Integer is Expressible as Product of Primes there exists $p \in \Z$ such that $p$ is prime and $p \divides k$.
That is:
:$\exists t \in \Z: k = p t$
and so:
:$m n = p^2 t^2 r s$
That is:
:$p^2 \divides m $
Hence the result by definition of the Möbius function.
Category:Multiplicative Functions
Category:Möbius Function
Möbius Inversion Formula

== Theorem ==
Let $f$ and $g$ be arithmetic functions.
Then:
:$(1): \quad \ds \map f n = \sum_{d \mathop \divides n} \map g d$
:
:$(2): \quad \ds \map g n = \sum_{d \mathop \divides n} \map f d \, \map \mu {\frac n d}$
where:
:$d \divides n$ denotes that $d$ is a divisor of $n$
:$\mu$ is the Möbius function.
=== Abelian Group ===
== Proof ==
Let $u$ be the unit arithmetic function and $\iota$ the identity arithmetic function.
Let $*$ denote Dirichlet convolution.
Then equation $(1)$ states that $f = g * u$ and $(2)$ states that $g = f * \mu$.
The proof rests on the following facts:
By Sum of Möbius Function over Divisors:
:$\mu * u = \iota$
By Properties of Dirichlet Convolution, Dirichlet convolution is commutative, associative and $h * \iota = h$ for all $h$.
We have:
Conversely:
Hence the result.
== Sources ==
* 
* 
Category:Dirichlet Convolution
Category:Möbius Function
Möbius Inversion Formula/Abelian Group

== Theorem ==
Let $G$ be an abelian group.
Let $f, g: \N \to G$ be mappings.
Then
:$\ds \map f n = \prod_{d \mathop \divides n} \map g d$
:
:$\ds \map g n = \prod_{d \mathop \divides n} \map f d^{\mu \paren {\frac n d} }$
where:
:$d \divides n$ denotes that $d$ is a divisor of $n$
:$\mu$ is the Möbius function.
== Proof ==
Category:Abelian Groups
Category:Möbius Function
Möbius Inversion Formula for Cyclotomic Polynomials

== Theorem ==
Let $n > 0$ be a (strictly) positive integer.
Let $\Phi_n$ be the $n$th cyclotomic polynomial.
Then:
:$\map {\Phi_n} x = \ds \prod_{d \mathop \divides n} \paren {x^d - 1}^{\map \mu {n / d} }$
where:
:the product runs over all divisors of $n$
:$\mu$ is the Möbius function.
== Proof ==
By Product of Cyclotomic Polynomials:
:$\ds \prod_{d \mathop \divides n} \map {\Phi_d} x = x^n - 1$
for all $n \in \N$.
The nonzero rational forms form an abelian group under multiplication.
By the Möbius inversion formula for abelian groups, this implies:
:$\ds \map {\Phi_n} x = \prod_{d \mathop \divides n} \paren {x^d - 1}^{\map \mu {n / d} }$
for all $n \in \N$.
Category:Cyclotomic Polynomials
Möbius Strip has Euler Characteristic Zero

== Theorem ==
Let $M$ be a Möbius Strip.
Then:
:$\map \chi M = 0$
where $\map \chi M$ denotes the Euler characteristic of the graph $M$.
== Proof ==
Let the number of vertices, edges and faces of $M$ be $V$, $E$ and $F$ respectively.
From Möbius Strip has no Vertices:
:$V = 0$
From Möbius Strip has 1 Edge:
:$E = 1$
From Möbius Strip has 1 Face:
:$F = 1$
By definition of the Euler characteristic:
== Source ==
* 
Category:Möbius Strip
Möbius Transformation is Bijection

== Theorem ==
Let $a, b, c, d \in \C$ be complex numbers.
Let $f: \overline \C \to \overline \C$ be the Möbius transformation:
:$\map f z = \begin {cases} \dfrac {a z + b} {c z + d} & : z \ne -\dfrac d c \\
\infty & : z = -\dfrac d c \\
\dfrac a c & : z = \infty \\
\infty & : z = \infty \text { and } c = 0 \end {cases}$
where $\overline \C$ denotes the extended complex plane.
Then:
:$f: \overline \C \to \overline \C$ is a bijection
:
:$a c - b d \ne 0$
=== Restriction to Reals ===
== Proof ==
We demonstrate that $f$ is injective $b c - a d \ne 0$.
demonstrating that when $z \ne -\dfrac d c$ and $z \ne \infty$:
:$\map f {z_1} = \map f {z_2} \implies z_1 = z_2$ $b c - a d \ne 0$
It remains to investigate the edge cases.
First we look at the case where $c \ne 0$.
That is, for $z \in \R$:
:$\dfrac {a z + b} {c z + d} = \dfrac c a$ only if $b c - a d = 0$
and so if $\map f {z_1} = \map f {z_2} = \dfrac a c$ it follows that $z_1 = z_2 = \infty$.
The case where $\map f {z_1} = \map f {z_2} = \infty$ follows by definition either that:
:$z_1 = z_2 = \dfrac a c$ when $c \ne 0$
or:
:$z_1 = z_2 = \infty$ when $c = 0$.
Thus we have that $f$ is an injection.
Now we investigate the inverse of $f$.
From Inverse Element of Injection we have that:
:$\map f z = w \implies \map {f^{-1} } w = z$
So, let $w = \map f z$.
First we recall that if $z = -\dfrac d c$, then $c z + d = 0$ and so $\dfrac {a z + b} {c z + d}$ is undefined.
Hence the need to investigate that case separately.
Take the general case, where $z \ne -\dfrac d c$ and $z \ne \infty$:
Thus we have that:
:$\map {f^{-1} } w = \dfrac {- d w + b} {c w - a}$
which is again a Möbius transformation, defined over all $w \in \C$ except where $w = \dfrac a c$.
We define:
:$\map {f^{-1} } {\dfrac a c} = \infty$
and:
:$\map {f^{-1} } \infty = -\dfrac d c$
except when $c = \infty$, where we define:
:$\map {f^{-1} } \infty = \infty$
Hence we have that the inverse of $f$ is another Möbius transformation.
So as $f^{-1}$ is also a Möbius transformation, it follows that:
:$\map {f^{-1} } {w_1} = \map {f^{-1} } {w_2} \implies w_1 = w_2$ $\paren {-d} c - b \paren {-a} = 0$
which is the same thing as $b c - a d \ne 0$.
Again, we have that $\dfrac {- d w + b} {c w - a} = -\dfrac d c$ only if $\paren {-d} c - b \paren {-a} = 0$.
As seen above, this is the same thing as $b c - a d \ne 0$.
Finally, we note that:
:$\map {f^{-1} } {w_1} = \map {f^{-1} } {w_2} = \infty \implies w_1 = w_2 = \dfrac {-d} c$
Thus we have that $f^{-1}$ is injective $b c - a d \ne 0$.
It follows from Injection is Bijection iff Inverse is Injection that $f$ is a bijection.
Category:Möbius Transformation is Bijection
Category:Möbius Transformations
Category:Examples of Bijections
Möbius Transformation is Bijection/Restriction to Reals

== Theorem ==
Let $a, b, c, d \in \R$ be real numbers.
Let $f: \R^* \to \R^*$ be the Möbius transformation restricted to the real numbers:
:$\map f x = \begin {cases} \dfrac {a x + b} {c x + d} & : x \ne -\dfrac d c \\
\infty & : x = -\dfrac d c \\
\dfrac a c & : x = \infty \\
\infty & : x = \infty \text { and } c = 0 \end {cases}$
Then:
:$f: \R^* \to \R^*$ is a bijection
:
:$a c - b d \ne 0$
== Proof ==
First we note that as Real Addition is Closed and Real Multiplication is Closed:
:$\Dom {\R^*} \subseteq \R^*$
Recall from Möbius Transformation is Bijection that the Möbius transformation on the extended complex plane is a bijection $a c - b d \ne 0$.
From Restriction of Injection is Injection, if $a c - b d \ne 0$ $f$ is an injection.
As the inverse $f^{-1}$ of $f$ is also the restriction of a Möbius transformation, it follows that $f^{-1}$ is also an injection.
Hence the result from Injection is Bijection iff Inverse is Injection.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions: Exercise $2$ 
Category:Möbius Transformation is Bijection
N-Tuple Root is Root of n-1th Derivatives of Equation but not nth Derivative

== Theorem ==
Let $\EE$ be the equation:
:$\map f x = 0$
where $f: \R \to \R$ is a real function.
Let $\xi$ be a multiple root of $\EE$ with a multiplicity of $n$.
Then for all $k \in \N: k < n$:
:$\xi$ is a root of the equation $\map {f^{\paren k} } x = 0$
:$\xi$ is not a root of the equation $\map {f^{\paren n} } x = 0$
where $f^{\paren k}: \R \to \R$ is the $k$th derivative of $f$ $x$.
== Proof ==
== Sources ==
* 
* 
Category:Multiple Roots
Category:Differential Calculus
NAND and NOR are Functionally Complete

== Theorem ==
The NAND and NOR operators are each functionally complete.
That is, NAND and NOR are Sheffer operators.
== Proof ==
In NAND is Functionally Complete it is demonstrated that NAND is functionally complete.
In NOR is Functionally Complete it is demonstrated that NOR is functionally complete.
== Historical Note ==
Category:Logical NAND
Category:Logical NOR
Category:Functional Completeness
NAND and NOR are Functionally Complete/Historical Note

== Historical Note on NAND and NOR are Functionally Complete ==
This result is specifically documented here because of its historical and scientific significance.
It was proved in $1913$ by , after whom the symbol $\uparrow$, the Sheffer stroke, is named after.
However, it later transpired that had noticed the same result in $1880$, but the paper in which this appeared was not published until $1933$.
Category:Historical Notes
NAND as Disjunction of Negations

== Theorem ==
:$p \uparrow q \dashv \vdash \neg p \lor \neg q$
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : $\S 2.5$: Further Logical Constants
Category:Logical NAND
Category:Disjunction
Category:NAND as Disjunction of Negations
NAND as Disjunction of Negations/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.
$\begin{array}{|ccc||ccccc|} \hline
p & \uparrow & q & \neg & p & \lor & \neg & q \\
\hline
\F & \T & \F & \T & \F & \T & \T & \F \\
\F & \T & \T & \T & \F & \T & \F & \T \\
\T & \T & \F & \F & \T & \T & \T & \F \\
\T & \F & \T & \F & \T & \F & \F & \T \\
\hline
\end{array}$
Category:NAND as Disjunction of Negations
Category:Truth Table Proofs
NAND is Commutative

== Theorem ==
Let $\uparrow$ signify the NAND operation.
Then, for any two propositions $p$ and $q$:
:$p \uparrow q \dashv \vdash q \uparrow p$
That is, NAND is commutative.
== Proof 1 ==
== Proof by Truth Table ==
== Also see ==
* NOR is Commutative
== Sources ==
* : $\S 2.3.3$
Category:Logical NAND
Category:NAND is Commutative
Category:Examples of Commutative Operations
NAND is Commutative/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables:
:$\begin{array}{|ccc||ccc|} \hline
p & \uparrow & q & q & \uparrow & p \\
\hline
\F & \T & \F & \F & \T & \F \\
\F & \T & \T & \T & \T & \F \\
\T & \T & \F & \F & \T & \T \\
\T & \F & \T & \T & \F & \T \\
\hline
\end{array}$
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
Category:NAND is Commutative
NAND is not Associative

== Theorem ==
Let $\uparrow$ signify the NAND operation.
Then there exist propositions $p, q, r$ such that:
:$p \uparrow \paren {q \uparrow r} \not \vdash \paren {p \uparrow q} \uparrow r$
That is, NAND is not associative.
== Proof 1 ==
== Proof by Truth Table ==
== Also see ==
* NOR is not Associative
== Sources ==
* : $\S 2.10$: Exercise $2.13$
Category:Logical NAND
Category:NAND is not Associative
NAND is not Associative/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables:
:$\begin{array}{|ccccc||ccccc|} \hline
p & \uparrow & (q & \uparrow & r) & (p & \uparrow & q) & \uparrow & r \\
\hline
\F & \T & \F & \T & \F & \F & \T & \F & \T & \F \\
\F & \T & \F & \T & \T & \F & \T & \F & \F & \T \\
\F & \T & \T & \T & \F & \F & \T & \T & \T & \F \\
\F & \T & \T & \F & \T & \F & \T & \T & \F & \T \\
\T & \F & \F & \T & \F & \T & \T & \F & \T & \F \\
\T & \F & \F & \T & \T & \T & \T & \F & \F & \T \\
\T & \F & \T & \T & \F & \T & \F & \T & \T & \F \\
\T & \T & \T & \F & \T & \T & \F & \T & \T & \T \\
\hline
\end{array}$
As can be seen by inspection, the truth values under the main connectives do not match for all boolean interpretations.
Category:NAND is not Associative
NAND with Equal Arguments

== Theorem ==
Let $\uparrow$ signify the NAND operation.
Then, for any proposition $p$:
:$p \uparrow p \dashv \vdash \neg p$
That is, the NAND of a proposition with itself corresponds to the negation operation.
== Proof 1 ==
== Proof by Truth Table ==
== Also see ==
* NOR with Equal Arguments
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(5)$
* : $\S 2.3.3$
Category:Logical NAND
Category:NAND with Equal Arguments
NAND with Equal Arguments/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables:
:$\begin{array}{|ccc||cc|} \hline
p & \uparrow & p & \neg & p \\
\hline
\F & \T & \F & \T & \F \\
\T & \F & \T & \F & \T \\
\hline
\end{array}$
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
== Sources ==
* : $\S 2.5$: Further Logical Constants
Category:NAND with Equal Arguments
Category:Truth Table Proofs
NOR is Commutative

== Theorem ==
Let $\downarrow$ signify the NOR operation.
Then, for any two propositions $p$ and $q$:
:$p \downarrow q \dashv \vdash q \downarrow p$
That is, NOR is commutative.
== Proof 1 ==
== Proof by Truth Table ==
== Also see ==
* NAND is Commutative
== Sources ==
* : $\S 2.3.3$
Category:Logical NOR
Category:NOR is Commutative
Category:Examples of Commutative Operations
NOR is Commutative/Proof by Truth Table

== Theorem ==
== Proof ==
Apply the Method of Truth Tables:
:$\begin{array}{|ccc||ccc|} \hline
p & \downarrow & q & q & \downarrow & p \\
\hline
\F & \T & \F & \F & \T & \F \\
\F & \F & \T & \T & \F & \F \\
\T & \F & \F & \F & \F & \T \\
\T & \F & \T & \T & \F & \T \\
\hline
\end{array}$
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
Category:NOR is Commutative