hcju/proofwikiqa huggingface dataset:
Let $x$ and $y$ be arbitrary. Let $a = d_{1'} \left({x_1, y_1}\right), b = d_{2'} \left({x_2, y_2}\right)$. {{WLOG}}, suppose that $\max \left\{ {a, b}\right\} = a$. Then: {{begin-eqn}} {{eqn | l = \lim_{p \mathop \to \infty} \left({a^p + b^p}\right)^{1/p}       | o = \ge       | r = \lim_{p \mathop \to \infty} \left({a^p}\right)^{1/p}       | c =  }} {{eqn | r = \lim_{p \mathop \to \infty} a       | c =  }} {{eqn | r = a       | c =  }} {{eqn | r = \max \left\{ {a, b}\right\}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \lim_{p \mathop \to \infty} \left({a^p + b^p}\right)^{1/p}       | o = \le       | r = \lim_{p \mathop \to \infty} \left({a^p + a^p}\right)^{1/p}       | c =  }} {{eqn | r = \lim_{p \mathop \to \infty} \left({2 a^p}\right)^{1/p}       | c =  }} {{eqn | r = \lim_{p \mathop \to \infty} a \cdot 2^{1/p}       | c =  }} {{eqn | r = a \lim_{p \mathop \to \infty} 2^{1/p}       | c =  }} {{eqn | r = a       | c =  }} {{eqn | r = \max \left\{ {a, b}\right\}       | c =  }} {{end-eqn}} The result follows by the Squeeze Theorem. {{qed}} Category:Chebyshev Distance Category:P-Product Metrics fsb3qf1xl03y3jluddwc0hl5tnkivtl
Suppose to the contrary that $\sqrt [3] 2$ is rational. Then: {{begin-eqn}} {{eqn | l = \sqrt [3] 2       | r = \frac p q       | c = for some integer $p$ and $q$ }} {{eqn | ll= \implies       | l = p^3       | r = q^3 + q^3       | c =  }} {{end-eqn}} which contradicts Fermat's Last Theorem. {{qed}} Category:Number Theory 8uo0v173rbvw5i3fyyqoqap1jy9fz41
$f$ satisfies the conditions of a [[Definition:Piecewise Continuously Differentiable Function/Definition 2|Piecewise Continuously Differentiable  Function by Definition 2]]. Therefore, there exists a finite subdivision $\set {x_0, \ldots, x_n}$ of $\closedint a b$, $x_0 = a$ and $x_n = b$, such that $f$ is continuously differentiable on $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$. Here, the derivatives at $x_{i - 1}$ and $x_i$ are understood as one-sided derivatives. $f$ is differentiable on $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$. Therefore, by Differentiable Function is Continuous, $f$ is continuous at every point of $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$. Here, the continuities at $x_{i - 1}$ and $x_i$ are one-sided. We use this result to go through every point of every $\closedint {x_{i - 1} } {x_i}$ in order to establish the continuity of $f$ there: $f$ is continuous on $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$. $\openint {x_{i - 1} } {x_i}$ is a subset of $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$. Therefore, $f$ is continuous at every point of $\openint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$. $x_0$ ($= a$) is the leftmost point in the domain of $f$. $f$ is right-continuous at $x_0$. Therefore, $f$ is continuous at $x_0$. $x_n$ ($= b$) is the rightmost point in the domain of $f$. $f$ is left-continuous at $x_n$. Therefore, $f$ is continuous at $x_n$. $f$ is continuous at every point $x_i$, $i \in \set {1, \ldots, n - 1}$, because $f$ is both left- and right-continuous at these points. All in all, we have found that $f$ is continuous at every point of $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$. $f$ is therefore continuous at every point of the union of those intervals: $\displaystyle \bigcup_{i \mathop = 1}^n \closedint {x_{i - 1}} {x_i}$. $\displaystyle \bigcup_{i \mathop = 1}^n \closedint {x_{i - 1}} {x_i}$ equals $\closedint a b$, the domain of $f$. Therefore, $f$ is continuous as $f$ is continuous at every point of its domain. {{qed}} Category:Real Analysis 4dnzpl7tuw8rcwjsdhf6kem98mzi2bk
=== Definition 2 implies Definition 1 === Assume that $f$ satisfies the requirement of Definition 2. We need to prove that $f$ satisfies the requirements of Definition 1. $f$ is continuous by Piecewise Continuously Differentiable Function/Definition 2 is Continuous. This proves $(1)$ in Definition 1. $f$ is continuously differentiable on $\closedint {x_{i - 1} } {x_i}$, the derivatives at $x_{i - 1}$ and $x_i$ understood as one-sided derivatives, for every $i \in \set {1, \ldots, n}$, by definition. Therefore, $f$ is continuously differentiable at every point of every $\openint {x_{i - 1} } {x_i}$ since $\openint {x_{i - 1} } {x_i}$ is a subset of $\closedint {x_{i - 1} } {x_i}$. This proves $(2.1)$ in Definition 1. $f$ has a continuous right-hand derivative at $x_{i - 1}$ for every $i \in \set {1, \ldots, n}$. This means that: :$\map {f'_+} {x_{i - 1} } = \displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} \map {f'} x$ This implies that $\displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} \map {f'} x$ exists for every $i \in \set {1, \ldots, n}$. Also, $f$ has a continuous left-hand derivative at $x_i$ for every $i \in \set {1, \ldots, n}$. This means that :$\map {f'_-} {x_i} = \displaystyle \lim_{x \mathop \to {x_i}^-} \map {f'} x$ This implies that $\displaystyle \lim_{x \mathop \to {x_i}^-} \map {f'} x$ exists for every $i \in \set {1, \ldots, n}$. These two conclusions prove that requirement $(2.2)$ in Definition 1 is satisfied. This finishes the proof that $f$ satisfies the requirements of Definition 1. {{qed|lemma}} === Definition 1 implies Definition 2 === Assume that $f$ satisfies the requirements of Definition 1. We need to prove that $f$ satisfies the requirement of Definition 2. $f$ is continuously differentiable on $\openint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$ by definition. Also, the one-sided limits $\displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} \map {f'} x$ and $\displaystyle \lim_{x \mathop \to {x_i}^-} \map {f'} x$ exist for every $i \in \set {1, \ldots, n}$. $f$ is continuously differentiable on $\closedint {x_{i - 1} } {x_i}$, the derivatives at $x_{i - 1}$ and $x_i$ understood as one-sided derivatives, for every $i \in \set {1, \ldots, n}$, by Extendability Theorem for Derivatives Continuous on Open Intervals. This is exactly the requirement of Definition 2. {{qed}} Category:Piecewise Continuously Differentiable Functions rbncfcw1bu62mjlmgotfd7i7zm49mkm
:300px Let $ABCD$ be a square inscribed in a circle. Let $EFGH$ be a square circumscribed around the same circle. We have that: :$ABCD$ is twice the area of the triangle $ADB$. :$EFGH$ is twice the area of the rectangle $EFBD$. From Area of Rectangle, the area of $EFBD$ is $ED \cdot DB$. From Area of Triangle in Terms of Side and Altitude, the area of $ADB$ is $\dfrac 1 2 \cdot ED \cdot DB$. Thus: :$EFGH = 2 \cdot ABCD$ But the area of $EFGH$ is greater than the area of the circle around which it is circumscribed. Therefore half of the area of $EFGH$ is greater than half of the area of the circle around which it is circumscribed. Therefore the area of $ABCD$ is greater than half of the area of the circle within which it is inscribed. {{qed}}
We are given that $f$ is piecewise continuous and bounded on $\closedint a b$. Therefore, there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that for all $i \in \set {1, 2, \ldots, n}$: :$f$ is continuous on $\openint {x_{i - 1} } {x_i}$ :$f$ is bounded on $\closedint {x_{i - 1} } {x_i}$. Note that $n$ is the number of intervals $\openint {x_{i - 1} } {x_i}$ defined from the (finite) subdivision $\set {x_0, x_1, \ldots, x_n}$. We shall use proof by induction on these $n$ intervals. For all $k \in \set {1, 2, \ldots, n}$, let $\map P k$ be the proposition: :$f$ is Darboux integrable on $\closedint {x_0} {x_k}$. === Basis for the Induction === $\map P 1$ is the case: :$f$ is Darboux integrable on $\closedint {x_{i - 1} } {x_i}$ for an arbitrary $i \in \set {1, 2, \ldots, k}$.  $f$ is piecewise continuous and bounded for the case $n = 1$ means that: :$f$ is continuous on $\openint {x_{i - 1} } {x_i}$ :$f$ is bounded on $\closedint {x_{i - 1} } {x_i}$. By Bounded Function Continuous on Open Interval is Darboux Integrable, $f$ is Darboux integrable on $\closedint {x_{i - 1} } {x_i}$. Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown 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 the induction hypothesis: :$f$ is Darboux integrable on $\closedint {x_0} {x_k}$ from which it is to be shown that: :$f$ is Darboux integrable on $\closedint {x_0} {x_{k + 1} }$. === Induction Step === This is the induction step: By definition of a bounded piecewise continuous function, for every $i \in \set {1, 2, \ldots, k, k + 1}$: :$f$ is continuous on $\openint {x_{i - 1} } {x_i}$ :$f$ is bounded on $\closedint {x_{i - 1} } {x_i}$. By the induction hypothesis, $f$ is Darboux integrable on $\closedint {x_0} {x_k}$. From the basis for the induction, $f$ is Darboux integrable on $\closedint {x_k} {x_{k + 1} }$. We have that $f$ is Darboux integrable on $\closedint {x_0} {x_k}$ and $\closedint {x_k} {x_{k + 1} }$. Therefore, $f$ is Darboux integrable on $\closedint {x_0} {x_k} \cup \closedint {x_k} {x_{k + 1} }$ by Existence of Integral on Union of Adjacent Intervals. We have that: :$\closedint {x_0} {x_{k + 1} } = \closedint {x_0} {x_k} \cup \closedint {x_k} {x_{k + 1} }$. Accordingly, $f$ is Darboux integrable on $\closedint {x_0} {x_{k + 1} }$. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$f$ is Darboux integrable on $\closedint a b$. {{qed}}
Consider the function: :$f \left({x}\right) = \begin{cases} 0 & : x = a \\ \sin \left({\dfrac 1 {x - a} }\right) & : x \in \left({a \,.\,.\, b}\right] \end{cases}$ Consider the (finite) subdivision $\left\{{a, b}\right\}$ of $\left[{a \,.\,.\, b}\right]$. We observe that $\sin \left({\dfrac 1 {x - a} }\right)$ is continuous on $\left({a \,.\,.\, b}\right)$. Since $f \left({x}\right) = \sin \left({\dfrac 1 {x - a} }\right)$ on $\left({a \,.\,.\, b}\right)$, it follows that $f$ is continuous on $\left({a \,.\,.\, b}\right)$. Also, $f$ is bounded by the bound $1$ on $\left[{a \,.\,.\, b}\right]$. Therefore $f$ is a bounded piecewise continuous function on the closed interval $\left[{a \,.\,.\, b}\right]$. We now investigate whether $f$ has one-sided limits at the endpoints of $\left[{a \,.\,.\, b}\right]$. The function $\sin \left({\dfrac 1 {x - a} }\right)$ varies between $-1$ and $+1$ as $x$ approaches $a$ from above. Thus it does not converge.  Since $f \left({x}\right) = \sin \left({\dfrac 1 {x - a} }\right)$ when $x > a$ we conclude that $\displaystyle \lim_{x \mathop \to a^+} f \left({x}\right)$ does not exist either. $f$ has no one-sided limit at $a$. Hence $f$ is not a piecewise continuous function with one-sided limits. {{qed}} Category:Piecewise Continuous Functions kkhpxp4jlyfs9es6kwm8a87h9dy5z1y
From {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :the square on a medial, applied to a rational straight line, produces as breadth a rational straight line which is incommensurable in length with that to which it is applied. From {{EuclidPropLink|book = X|prop = 60|title = Square on Binomial Straight Line applied to Rational Straight Line}}: :the square on a binomial, applied to a rational straight line, produces as breadth the first binomial. From {{EuclidPropLink|book = X|prop = 61|title = Square on First Bimedial Straight Line applied to Rational Straight Line}}: :the square on a first bimedial, applied to a rational straight line, produces as breadth the second binomial. From {{EuclidPropLink|book = X|prop = 62|title = Square on Second Bimedial Straight Line applied to Rational Straight Line}}: :the square on a second bimedial, applied to a rational straight line, produces as breadth the third binomial. From {{EuclidPropLink|book = X|prop = 63|title = Square on Major Straight Line applied to Rational Straight Line}}: :the square on the major, applied to a rational straight line, produces as breadth the fourth binomial. From {{EuclidPropLink|book = X|prop = 64|title = Square on Side of Rational plus Medial Area applied to Rational Straight Line}}: :the square on the side of a rational plus medial area, applied to a rational straight line, produces as breadth the fifth binomial. From {{EuclidPropLink|book = X|prop = 65|title = Square on Side of Sum of two Medial Area applied to Rational Straight Line}}: :the square on the side of the sum of two medial areas, applied to a rational straight line, produces as breadth the sixth binomial. All of these breadths so produced differ from the first and from each other: :from the first because it is rational and: :from each other because they are different in order. {{qed}} {{Euclid Note|72|X}}
From {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :the square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied. From {{EuclidPropLink|book = X|prop = 97|title = Square on Apotome applied to Rational Straight Line}}: :the square on an apotome applied to a rational straight line produces as breadth a first apotome. From {{EuclidPropLink|book = X|prop = 98|title = Square on First Apotome of Medial Straight Line applied to Rational Straight Line}}: :the square on a first apotome of a medial straight line applied to a rational straight line produces as breadth a second apotome. From {{EuclidPropLink|book = X|prop = 99|title = Square on Second Apotome of Medial Straight Line applied to Rational Straight Line}}: :the square on a second apotome of a medial straight line applied to a rational straight line produces as breadth a third apotome. From {{EuclidPropLink|book = X|prop = 100|title = Square on Minor Straight Line applied to Rational Straight Line}}: :the square on a minor straight line applied to a rational straight line produces as breadth a fourth apotome. From {{EuclidPropLink|book = X|prop = 101|title = Square on Straight Line which produces Medial Whole with Rational Area applied to Rational Straight Line}}: :the square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome. From {{EuclidPropLink|book = X|prop = 102|title = Square on Straight Line which produces Medial Whole with Medial Area applied to Rational Straight Line}}: :the square on the straight line which produces with a medial area a medial whole, if applied to a rational straight line, produces as breadth a sixth apotome. All of these breadths so produced differ from the first and from each other: :from the first because it is rational and: :from each other because they are different in order. Thus it follows that the irrational straight lines themselves are different from one another. From {{EuclidPropLink|book = X|prop = 111|title = Apotome not same with Binomial Straight Line}}: :an apotome is not the same as a binomial straight line. Combining the above analysis with Classification of Irrational Straight Lines derived from Binomial Straight Line, it can be seen that the following $13$ types of irrational straight line are all different in nature: :Medial :Binomial :First bimedial :Second bimedial :Major :"Side" of a rational plus medial area :"Side" of the sum of two medial areas :Apotome :First apotome of a medial straight line :Second apotome of a medial straight line :Minor :Straight line which produces with a rational area a medial whole :Straight line which produces with a medial area a medial whole. {{qed}} {{Euclid Note|111|X}}
Let $x \in \Q$, where $\Q$ denotes the set of rational numbers. Consider the rational sequence: :$x, x, x, \ldots$ This sequence is trivially Cauchy. Thus there exists a Cauchy sequence $\eqclass {\sequence {x_n} } {}$ which is identified with a rational number $x \in \Q$ such that: So by the definition of a real number: :$x \in \R$ where $\R$ denotes the set of real numbers. Thus, by definition of subset: :$\Q \subseteq \R$ {{qed}}
By Proof by Counterexample: Let $x = \sqrt 2$. From Square Root of 2 is Irrational: :$\sqrt 2$ is an irrational number. By definition: :$x \in \R \setminus \Q$ where: :$\R$ is the set of real numbers :$\Q$ is the set of rational numbers :$\setminus$ denotes the set difference. Thus $x$, while being a real number, is not also a rational number. Hence the result. {{qed}}
Let $x$ be a real number which is both positive and negative. Thus: :$x \in \set {x \in \R: x \ge 0}$ and: :$x \in \set {x \in \R: x \le 0}$ and so: :$0 \le x \le 0$ from which: :$x = 0$ {{qed}}
Let $\Bbb U$ be a universal set. Let $\circ: S \times S \to \Bbb U$ be a binary operation on $S$. Let $T \subseteq S$. Let $\left({a, b}\right) \in T \times T$. By definition of ordered pair and cartesian product: :$a \in T$ and $b \in T$ As $T \subseteq S$, it follows that: :$a \in S$ and $b \in S$ Thus: :$\left({a, b}\right) \in S \times S$ As $\circ$ is a binary operation on $S$, it follows that: :$\circ \left({a, b}\right) \in \Bbb U$ But by definition of restriction of $\circ$: :$\circ \left({a, b}\right) = \circ {\restriction}_T \left({a, b}\right)$ Thus: :$\circ {\restriction}_T \left({a, b}\right) \in \Bbb U$ As $a$ and $b$ are arbitrary elements of $T$, it follows that this holds for all $a, b \in T$. Hence the result by definition of binary operation. {{qed}} Category:Operations owksju1jdr9447lowaf79mvn8kii969
Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{m n}$ and $\mathbf C = \sqbrk c_{m n}$ be elements of the $m \times n$ matrix space over $R$. Then: {{begin-eqn}} {{eqn | l = \paren {\mathbf A + \mathbf B} + \mathbf C       | r = \paren {\sqbrk a_{m n} + \sqbrk b_{m n} } + \sqbrk c_{m n}       | c = Definition of $\mathbf A$, $\mathbf B$ and $\mathbf C$ }} {{eqn | r = \sqbrk {a + b}_{m n} + \sqbrk c_{m n}       | c = {{Defof|Matrix Entrywise Addition over Ring}} }} {{eqn | r = \sqbrk {\paren {a + b} + c}_{m n}       | c = {{Defof|Matrix Entrywise Addition over Ring}} }} {{eqn | r = \sqbrk {a + \paren {b + c} }_{m n}       | c = {{Ring-axiom|A1}} }} {{eqn | r = \sqbrk a_{m n} + \sqbrk {b + c}_{m n}       | c = {{Defof|Matrix Entrywise Addition over Ring}} }} {{eqn | r = \sqbrk a_{m n} + \paren {\sqbrk b_{m n} + \sqbrk c_{m n} }       | c = {{Defof|Matrix Entrywise Addition over Ring}} }} {{eqn | r = \mathbf A + \paren {\mathbf B + \mathbf C}       | c = Definition of $\mathbf A$, $\mathbf B$ and $\mathbf C$ }} {{end-eqn}} {{qed}}
By definition, matrix entrywise addition is the '''Hadamard product''' of $\mathbf A$ and $\mathbf B$ with respect to ring addition. We have from {{Ring-axiom|A1}} that ring addition is associative. The result then follows directly from Associativity of Hadamard Product. {{qed}}
By definition of composition of mappings: :$\phi \circ \psi$ is a mapping. We have by definition of symmetry that: :$\map \phi P$ is congruent to $P$ and: :$\map \psi {\map \phi P}$ is congruent to $\map \phi P$ Therefore: :$\phi \circ \psi$ is congruent to $P$ Thus $\phi \circ \psi$ is a symmetry of $P$. {{qed}}
From Composition of Symmetries is Symmetry: :$\paren {\phi \circ \psi} \circ \chi$ is a symmetry and: :$\phi \circ \paren {\psi \circ \chi}$ is a symmetry. It follows from Composition of Mappings is Associative that: : $\paren {\phi \circ \psi} \circ \chi = \phi \circ \paren {\psi \circ \chi}$ {{qed}}
We have that $1 \in \Z_m$ and that: :$1 \cdot_m 1 = 1$ We have that $m - 1 \in \Z_m$ and that: :$\left({m - 1}\right) \cdot_m \left({m - 1}\right) = 1$ Thus unless $m - 1 = 1$, that is, $m = 2$, there exist $2$ residues of $\Z_m$ whose product modulo $m$ with itself equals $1$. There are $m - 2$ residues which, when multiplied modulo $m$ with themselves have as a result a residue. Thus there can be at maximum $m - 2$ residues (excluding $1$) which can be the product modulo $m$ of a residue with itself. But there are $m - 1$ residues (excluding $1$). So at least $1$ residues is not the product modulo $m$ of a residue with itself. {{qed}}
{{begin-eqn}} {{eqn | l = \relcomp S {\bigcap \mathbb T}       | r = S \setminus \paren {\bigcap \mathbb T}       | c = {{Defof|Relative Complement}} }} {{eqn | r = \bigcup_{H \mathop \in \mathbb T} \paren {S \setminus H}       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \bigcup_{H \mathop \in \mathbb T} \relcomp S H       | c = {{Defof|Relative Complement}} }} {{end-eqn}} {{qed}} {{Namedfor|Augustus De Morgan}}
{{begin-eqn}} {{eqn | l = \relcomp S {\bigcup \mathbb T}       | r = S \setminus \paren {\bigcup \mathbb T}       | c = {{Defof|Relative Complement}} }} {{eqn | r = \bigcap_{H \mathop \in \mathbb T} \paren {S \setminus H}       | c = De Morgan's Laws: Difference with Union }} {{eqn | r = \bigcap_{H \mathop \in \mathbb T} \relcomp S H       | c = {{Defof|Relative Complement}} }} {{end-eqn}} {{qed}} {{Namedfor|Augustus De Morgan}}
{{begin-eqn}} {{eqn | l = \relcomp S {\bigcup_{i \mathop \in I} S_i}       | r = S \setminus \paren {\bigcup_{i \mathop \in I} S_i}       | c = {{Defof|Relative Complement}} }} {{eqn | r = \bigcap_{i \mathop \in I} \paren {S \setminus S_i}       | c = De Morgan's Laws for Set Difference: Difference with Union }} {{eqn | r = \bigcap_{i \mathop \in I} \relcomp S {S_i}       | c = {{Defof|Relative Complement}} }} {{end-eqn}} {{qed}} {{Namedfor|Augustus De Morgan}}
{{begin-eqn}} {{eqn | l = \relcomp S {\bigcap_{i \mathop \in I} S_i}       | r = S \setminus \paren {\bigcap_{i \mathop \in I} S_i}       | c = {{Defof|Relative Complement}} }} {{eqn | r = \bigcup_{i \mathop \in I} \paren {S \setminus S_i}       | c = De Morgan's Laws for Set Difference: Difference with Intersection }} {{eqn | r = \bigcup_{i \mathop \in I} \relcomp S {S_i}       | c = {{Defof|Relative Complement}} }} {{end-eqn}} {{qed}} {{Namedfor|Augustus De Morgan}}
Let $T_1, T_2 \subseteq S$. Then from Union is Smallest Superset: :$T_1 \cup T_2 \subseteq S$ Hence: {{begin-eqn}} {{eqn | l = \relcomp S {T_1 \cup T_2}       | r = S \setminus \paren {T_1 \cup T_2}       | c = {{Defof|Relative Complement}} }} {{eqn | r = \paren {S \setminus T_1} \cap \paren {S \setminus T_2}       | c = De Morgan's Laws: Difference with Union }} {{eqn | r = \relcomp S {T_1} \cap \relcomp S {T_2}       | c = {{Defof|Relative Complement}} }} {{end-eqn}} {{qed}} {{Namedfor|Augustus De Morgan}}
Let $T_1, T_2 \subseteq S$. Then from Intersection is Subset and Subset Relation is Transitive: :$T_1 \cap T_2 \subseteq S$ Hence: {{begin-eqn}} {{eqn | l = \relcomp S {T_1 \cap T_2}       | r = S \setminus \paren {T_1 \cap T_2}       | c = {{Defof|Relative Complement}} }} {{eqn | r = \paren {S \setminus T_1} \cup \paren {S \setminus T_2}       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \relcomp S {T_1} \cup \relcomp S {T_2}       | c = {{Defof|Relative Complement}} }} {{end-eqn}} {{qed}} {{Namedfor|Augustus De Morgan}}
Suppose: :$\displaystyle x \in S \setminus \bigcap_{i \mathop \in I} T_i$ Note that by Set Difference is Subset we have that $x \in S$ (we need this later). Then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = S \setminus \bigcap_{i \mathop \in I} T_i       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \notin       | r = \bigcap_{i \mathop \in I} T_i       | c = {{Defof|Set Difference}} }} {{eqn | ll= \leadstoandfrom       | l = \neg \leftparen {\forall i \in I: x}       | o = \in       | r = \rightparen {T_i}       | c = {{Defof|Intersection of Family}} }} {{eqn | ll= \leadstoandfrom       | l = \exists i \in I: \neg \leftparen x       | o = \in       | r = \rightparen {T_i}       | c = De Morgan's Laws (Predicate Logic) }} {{eqn | ll= \leadstoandfrom       | l = \exists i \in I: x       | o = \in       | r = S \setminus T_i       | c = {{Defof|Set Difference}}: note $x \in S$ from above }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}       | c = {{Defof|Union of Family}} }} {{end-eqn}} Therefore: :$\displaystyle S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$ {{qed}} {{Namedfor|Augustus De Morgan}}