hcju/proofwikiqa huggingface dataset:
Let $P$ be the property defined as: :$\forall x: \map P x := \paren {x \subseteq a}$ where $\lor$ is the disjunction operator. Hence we form the set: :$\powerset a := \set {x: x \subseteq a}$ {{qed}}
Let $P$ be the property defined as: :$\forall x: \map P x := \paren {\exists y: y \in a \land x \in y}$ where $\land$ is the conjunction operator. That is, $\map P x$ {{iff}}: :$x$ is an element of some set $y$, where $y$ is one of the sets that comprise the elements of $a$. Hence we form the set: :$\bigcup a := \set {x: \exists y: y \in a \land x \in y}$ {{qed}}
From Russell's Paradox, the comprehension principle leads to a contradiction. Let $q$ be such a contradiction: :$q = p \land \neg p$ for some statement $p$. From the Rule of Explosion it then follows that every logical formula is a provable consequence of $q$. Hence the result, by definition of inconsistent. {{qed}}
Let $S$ be a set. Let $T$ be the set of all elements of $S$ which do not contain $S$ as elements. Then by the corollary to Russell's paradox $T$ itself cannot be an element of $S$. This $T$ is the required subset.
Consider the power set $\powerset S$ of $S$. From Cantor's Theorem, there is no surjection $f: S \to \powerset S$. That is, there are more subsets of $S$ than there are elements of $S$. So there must be at least one subset of $S$ which is not an element of $S$. {{qed}}
By the definition of the box topology: :$V \in \tau_b$ Let $W \in \tau_T$. === Case 1 === Let $W \cap V = \O$. Since $U \neq \O$, there exists $x \in U$. Then $\family {x}_{n \in \N} \in V$ So $\family {x}_{n \in \N} \notin W$ By the definition of set equality: :$V \neq W$ {{qed|lemma}} === Case 2 === Let $W \cap V \neq \O$. From Natural Basis of Tychonoff Topology, $W = \displaystyle \prod_{n \mathop \in \N } W_n$ where: :for all $n \in \N : W_n \in \tau$ :for all but finitely many indices $n : W_n = X$ Let $m = \max \set{n : W_n \neq X}$ Then: :$W_{m + 1} = X$ Since $W \cap V \neq \O$, there exists $\family {z_n} \in W \cap V$. Since $U \neq X$, there exists $y \in X \setminus U$. Let $\family {y_n}$ be defined as: :$y_n = \begin{cases} z_n & : n \ne m + 1 \\ y  & : n = m + 1 \end{cases}$ Since $y_{m + 1} = y \in X = W_{m + 1}$, it follows that: : $\family {y_n} \in W$ Since $y_{m + 1} = y \notin U$, it follows that: : $\family {y_n} \notin V$ By the definition of set equality: :$W \neq V$ {{qed|lemma}} In either case $W \neq V$. Since $W$ was arbitrary, it follows that for all $W \in \tau_T$, $W \neq V$. Thus $V \notin \tau_T$   {{qed}} Category:Box Topology c53k3bvylo1nbzjj2v32kimplgq4tvg
From P-adic Number times P-adic Norm is P-adic Unit there exists $n \in \Z$: :$\mathbf p^n \mathbf a \in \Z^\times_p$ :$\norm {\mathbf a}_p =  p^n$  where $\Z^\times_p$ is the set of $p$-adic units. Let $l = -n$. From P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient: :$l = \min \set {i: i \ge m \land d_i \ne 0}$ and :$l = \min \set {i: i \ge k \land e_i \ne 0}$ Now: {{begin-eqn}} {{eqn | l = m       | o = <       | r = 0 }} {{eqn | ll= \leadsto       | l = d_m       | o = \ne       | r = 0       | c = {{Defof|P-adic Expansion|$p$-adic Expansion}} }} {{eqn | ll= \leadsto       | l = m       | r = l       | c = Definition of $l$ }} {{eqn | ll= \leadsto       | l = l       | o = <       | r = 0 }} {{end-eqn}} Similarly: :$k < 0 \leadsto k = l$ === Proof of statement $(1)$ === ==== Case $l < 0$ ==== Let $l < 0$. Then: {{begin-eqn}} {{eqn | l = l       | o = <       | r = 0 }} {{eqn | ll= \leadsto       | l = m       | o = <       | r = 0       | c = as $m \le l$ }} {{eqn | ll= \leadsto       | l = m       | r = l       | c = from above }} {{end-eqn}} Similarly: :$l < 0 \leadsto k = l$ So:  :$m = l = k$. {{qed|lemma}} ==== Case $l \ge 0$ ==== Let $l \ge 0$. Then: {{begin-eqn}} {{eqn | l = l       | o = \ge       | r = 0 }} {{eqn | ll= \leadsto       | l = m       | o = \not <       | r = 0       | c = as $m < 0 \leadsto l < 0$ }} {{eqn | ll= \leadsto       | l = m       | r = 0       | c = {{Defof|P-adic Expansion|$p$-adic Expansion}} }} {{end-eqn}} Similarly: :$l \ge  0 \leadsto k = 0$ So: :$m = 0 = k$. {{qed|lemma}} === Proof of statement $(2)$ === By definition of $l$, for all $i$ such that $m \le i < l$: :$d_i = 0$ :$e_i = 0$ So for all $m \le i < l$: :$d_i = e_i$. Consider the series: :$\displaystyle \sum_{i \mathop = l}^\infty d_i p^i$ From P-adic Expansion Less Intial Zero Terms Represents Same P-adic Number :$\displaystyle \sum_{i \mathop = l}^\infty d_i p^i$ is a representative of $\mathbf a$ From Multiple Rule for Cauchy Sequences in Normed Division Ring: :$\displaystyle p^{-l} \sum_{i \mathop = l}^\infty d_i p^i = \sum_{i \mathop = l}^\infty d_i p^{i - l} = \sum_{i \mathop = 0}^\infty d_{i + l} p^i$ is a representative of $\mathbf p^{-l} \mathbf a \in \Z^\times_p$. By definition of a $p$-adic expansion: :$\forall i \in N : 0 \le d_{i + 1} < p - 1$ Thus the series: :$\displaystyle \sum_{i \mathop = 0}^\infty d_{i + l} p^i$ is a $p$-adic expansion that represents $\mathbf p^{-l} \mathbf a \in \Z^\times_p$. Similarly the series: :$\displaystyle \sum_{i \mathop = 0}^\infty e_{i + l} p^i$ is a $p$-adic expansion that represents $\mathbf p^{-l} \mathbf a \in \Z^\times_p$. From Equivalence Class in P-adic Integers Contains Unique P-adic Expansion it follows that: :$\forall i \in N : d_{i + 1} = e_{i + 1}$ That is: :$\forall i \ge l : d_i = e_i$ The result follows. {{qed}} Category:P-adic Number Theory 2cnda3u7x2q2m3rcclpg44x05lx9hob
By definition of class, $A$ is a collection of sets. Let $x \in A$ be a set. By definition of universal class, $V$ contains all sets as elements. Hence $x \in V$. So we have that: :$x \in A \implies x \in V$ and the result follows by definition of subclass. {{qed}}
Let a set $x$ be defined as '''ordinary''' {{iff}} $x \notin x$. Let $\map \phi x$ be the set property defined as: :$\map \pi x := \neg \paren {x \in x}$ Then by the axiom of specification there exists a class, which can be denoted $A$, such that: :$A = \set {x: \neg \paren {x \in x} }$ By the axiom of extension $A$ is unique. Thus $A$ is the class of all '''ordinary''' sets. Hence we have: :$x \in A \iff x \notin x$ {{AimForCont}} $A$ were a set. Then we could set $A$ for $x$, and so obtain: :$A \in A \iff A \notin A$ This is a contradiction. Hence by Proof by Contradiction $A$ cannot be a set. But as $A$ is a subclass of the universal class $V$, $A$ is a class. So the class $A$ of '''ordinary''' sets is not a set. That is: :$A \notin V$ {{qed}}
Let a set $x$ be defined as '''ordinary''' {{iff}} $x \notin x$. Let $\map \phi x$ be the set property defined as: :$\map \pi x := \neg \paren {x \in x}$ Then by the axiom of specification there exists a class, which can be denoted $B$, such that: :$B = \set {x \in A: \neg \paren {x \in x} }$ That is, $B$ contains as elements of all the '''ordinary''' sets of $A$. By the axiom of extension $B$ is unique. Thus $B$ is the class of all '''ordinary''' sets of $A$. :$x \in B \iff \paren {x \in A \land x \notin x}$ Hence, for all $x$ which happen to be elements of $A$, we have: :$x \in B \iff x \notin x$ {{AimForCont}} $B \in A$. Then we could set $A$ for $x$, and so obtain: :$B \in B \iff B \notin B$ This is a contradiction. Hence by Proof by Contradiction $B$ cannot be an element of $A$. That is: :$B \notin V$ {{qed}}
By definition, a class $V$ is supercomplete {{iff}} $V$ is both transitive and swelled. From the axiom of transitivity, $V$ is transitive. From the axiom of swelledness, $V$ is swelled. Hence the result. {{qed}}
{{AimForCont}} $V$ were a set. Then by the axiom of swelledness, $V$ is swelled. That is, as $V$ is a set, every subclass of $V$ would also be a set. From Class has Subclass which is not Element, $V$ has a subclass $S$ which is not an element of $V$. But by definition of a basic universe, $V$ is a universal class. That is, $S \in V$.  This contradicts the deduction that $S \notin V$. Hence the result by Proof by Contradiction. {{qed}}
Let $P$ be a property such that $\map P x$ is satisfied by no $x$ at all, for example: :$\forall x: \map P x := \neg {x = x}$ Then by the axiom of specification we can create the class $A$ such that: :$A := \set {x \in V \land \neg {x = x} }$ from which it is seen that $A$ has no elements. Hence there exists an empty class. Let $A$ and $B$ both be empty classes. By definition, both $A$ and $B$ contain the same elements, that is, no elements at all. By the axiom of extension, that means $A = B$. Hence the result. {{qed}}
Let $A$ be a class. By definition of the empty class: :$\forall x: \neg \paren {x \in \O}$ From False Statement implies Every Statement: :$\forall x: \paren {x \in \O \implies x \in A}$ Hence the result by definition of subclass. {{qed}}
Vacuously, every element of $\O$ is also a subclass of $\O$. Hence $\O$ is transitive by definition. Vacuously, every subclass of every element of $\O$ is also an element of $\O$. Hence $\O$ is swelled by definition. The result follows by definition of supercomplete. {{qed}}
The axiom of the empty set gives us: {{:Axiom:Axiom of Empty Set/Class Theory}} Hence the result by definition of empty class. {{Qed}}
=== Necessary Condition === Let the axiom of the empty set hold. That is: :$\O$ is a set. By definition of a basic universe, $V$ is a universal class. Hence, by definition, every set is an element of $V$. We have that $\O$ is a set by hypothesis. Thus: :$\O \in V$ and by definition of empty class it is seen that $V$ is not empty. {{qed|lemma}} === Sufficient Condition === Let $V$ have at least one element. Let $x \in V$ be an element of $V$. Then by the axiom of transitivity $x$ is also a class. By Empty Class is Subclass of All Classes: :$\O \subseteq x$ By the axiom of swelledness, every subclass of $x$ is an element of $V$. That includes $\O$. Thus: :$\O \in V$ as we were to show. {{Qed}}
Using the axiom of specification, let $A$ be the class defined as: :$A := \set {x: x \in V \land x = a}$ That is: :$A = \set a$ By the axiom of extension, $\set a$ is the only such class which has $a$ as an element.
Let $x \in \set \O$ be any element of $\set \O$. Then it has to be the case that $x = \O$. Then every element of $\O$ is an element of $\set \O$ vacuously. That is, $\set \O$ is swelled. There is one element of $\set \O$, and that is $\O$. This is a subclass of $\set \O$. That is, $\set \O$ is transitive. The result follows by definition of supercomplete class. {{qed}}
Let $a = b$. Then $\set a$ and $\set b$ contain the same elements. Hence by the axiom of extension it follows that $\set a = \set b$. Let $\set a = \set b$. We have that $a \in \set a$. As $\set a = \set b$ we also have that $a \in \set b$. But $b$ is the only element of $\set b$. Hence it must be the case that $a = b$. {{qed}}
Using the axiom of specification, let $A$ be the class defined as: :$A := \set {x: x \in V \land \paren {x = a \lor x = b} }$ That is: :$A = \set {a, b}$ By the axiom of extension, $\set {a, b}$ is the only such class which has $a$ and $b$ as elements.
Let $A = \set {a, b}$ The existence of $A$ is shown in Doubleton Class can be Formed from Two Sets: :$A := \set {x: \paren {x = a \lor x = b} }$ Let $a = b$. Then: {{begin-eqn}} {{eqn | l = A       | r = \set {a, b}       | c = Definition of $A$ }} {{eqn | r = \set {a, a}       | c = as $a = b$ }} {{eqn | ll= \leadsto       | l = x \in \set {a, a}       | o = \iff       | r = \paren {x = a \lor x = a}       | c = Definition of $A$ }} {{eqn | o = \iff       | r = x = a       | c = Rule of Idempotence }} {{eqn | o = \iff       | r = x \in \set a       | c = {{Defof|Singleton Class}} }} {{end-eqn}} {{qed}}
=== $(1)$ implies $(2)$ === Let formulation $1$ of the axiom of pairing be assumed: :$\forall A: \forall B: \exists x: \forall y: \paren {y \in x \iff y = A \lor y = B}$ By definition of the biconditional, this can be expressed as: :$\forall A: \forall B: \exists x: \forall y: \paren {\paren {y \in x \implies y = A \lor y = B} \land \paren {y \in x \impliedby y = A \lor y = B} }$ from which, by the Rule of Simplification: :$\forall A: \forall B: \exists x: \forall y: \paren {y \in x \implies y = A \lor y = B}$ Thus formulation $2$ of the axiom of pairing is seen to hold. {{qed|lemma}} === $(2)$ implies $(1)$ === Let formulation $2$ of the axiom of pairing be assumed: :$\forall A: \forall B: \exists x: \forall y: \paren {y \in x \implies y = A \lor y = B}$ {{ProofWanted}} Thus formulation $1$ of the axiom of pairing is seen to hold. {{qed}} Category:Axiom of Pairing 68pi5r04te5ncpg0zcxdeznk6uc4npc
Let $x$ and $y$ be sets. Let $x = y$. From Doubleton Class of Equal Sets is Singleton Class, the doubleton class $\set {x, y}$  is the singleton class $\set x$. From the axiom of pairing, the doubleton class $\set {x, y}$ is a set when $x$ and $y$ are sets. Hence $\set x$ is a set. {{qed}}
Let $a_n$ be the class defined as: :$\forall n \in \N: a_n = \begin{cases} \O & : n = 0 \\ \set {a_{n - 1} } & : n > 0 \end {cases}$ It is shown by the Principle of Mathematical Induction that $a_n$ is a set for all $n \in \N$. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$a_n$ is a set. === Basis for the Induction === $\map P 0$ is the case: :$a_0$ is a set. But $a_0 = \O$ by definition. By the axiom of the empty set: :$\O$ is a set. :$\O \in V$ Thus $\map P 0$ 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: :$a_k$ is a set. from which it is to be shown that: :$a_{k + 1}$ is a set. === Induction Step === This is the induction step: By the induction hypothesis we have that $a_k$ is a set. From Singleton Class of Set is Set, its singleton class $\set {a_k}$ is also a set. But by definition: :$a_{k + 1} = \set {a_k}$ Hence $a_{k + 1}$ has been shown to be a set. So $\map P k \implies \map P {k + 1}$ and so by the Principle of Mathematical Induction, $a_n$ is a set for all $n \in \N$. {{qed|lemma}} It is now proved that $a_m \ne a_n$ when $m \ne n$. {{WLOG}}, let $m > n$. {{AimForCont}} $a_n = a_{m + n}$ for some $m > 0, n \ge 0$. Let $n = 0$. Then we would have: :$a_m = \O$ But as $m > 0$ we have that $a_m = \set {a_{m - 1} }$ for some $m - 1 \in \N$. But $\O \ne \set x$ for any set $x$. Now suppose $n > 0$. Then by Singleton Classes are Equal iff Sets are Equal: :$a_{n - 1} = a_{n - 1 + m}$ We can repeat this step a total of $n$ times to reach: :$a_0 = a_{0 + m}$ from which again it follows that $a_m = \O$ for some $m > 0$. This has been shown to be false for all $m$. Hence it cannot be the case that $a_n = a_{m + n}$ for some $m > 0, n \ge 0$. Thus there exists an injection from $\N$ to $\set {a_n: n \in \N}$. That is, $\set {a_n: n \in \N}$ is countably infinite by definition. By definition of basic universe, $V$ is a universal class. By definition of universal class, every set is an element of $V$. Hence: :$\forall n \in \N: a_n \in V$ and it follows that $V$ is infinite. {{qed}}
=== $(1)$ implies $(2)$ === Let formulation $1$ of the axiom of pairing be assumed: :The class $\set {a, b}$ is a set. Then we have that $c = \set {a, b}$ is a set such that both $a \in c$ and $b \in c$. Thus formulation $2$ of the axiom of pairing is seen to hold. {{qed|lemma}} === $(2)$ implies $(1)$ === Let formulation $2$ of the axiom of pairing be assumed: :There exists a set $c$ such that $a \in c$ and $b \in c$. Then the class $\set {a, b}$ is a subclass of $c$. By the axiom of swelledness, every subclass of $c$ is a set. That is, $\set {a, b}$ is aset. Thus formulation $1$ of the axiom of pairing is seen to hold. {{qed}}
We have that: :$b \in \set {a, b}$ and so by the axiom of extension: :$b \in \set {a, d}$ So: :$(1): \quad$ either $b = a$ or $b = d$. First suppose that $b = a$. Then: :$\set {a, b} = \set {a, a} = \set a$ We have that: :$d \in \set {a, d}$ and so by the axiom of extension: :$d \in \set {a, b}$ and so as $\set {a, b} = \set a$ it follows that: :$d = a$ We have $b = a$ and $d = a$ and so: :$b = d$ and so $b = d$. {{qed|lemma}} Next suppose that $b \ne a$. We have from $(1)$ that either $b = a$ or $b = d$. As $b \ne a$ it must be the case that $b = d$. {{qed|lemma}} So in either case we see that: :$b = d$ {{qed}}
Suppose $a = c$ and $b = d$. Then: :$\set a = \set c$ and: :$\set {a, b} = \set {c, d}$ Thus: :$\set {\set a, \set {a, b} } = \set {\set c, \set {c, d} }$ and so by the Kuratowski formalization: :$\tuple {a, b} = \tuple {c, d}$ {{qed}}
First a lemma: {{:Equality of Ordered Pairs/Lemma}}{{qed|lemma}} Let $\tuple {a, b} = \tuple {c, d}$. From the Kuratowski formalization: :$\set {\set a, \set {a, b} } = \set {\set c, \set {c, d} }$  There are two cases: either $a = b$, or $a \ne b$. ==== Case 1 ==== Suppose $a = b$. Then: :$\set {\set a, \set {a, b} } = \set {\set a, \set a} = \set {\set a}$ Thus $\set {\set c, \set {c, d} }$ has only one element. Thus $\set c = \set {c, d}$ and so $c = d$. So: : $\set {\set c, \set {c, d} } = \set {\set a}$ and so $a = c$ and $b = d$. Thus the result holds. {{qed|lemma}} ==== Case 2 ==== Now suppose $a \ne b$. By the same argument it follows that $c \ne d$. So that means that either $\set a = \set c$ or $\set a = \set {c, d}$. Since $\set {c, d}$ has $2$ distinct elements, $\set a \ne \set {c, d}$. Thus: :$\set a = \set c$ and so $a = c$. Then: :$\set {a, b} = \set {c, d}$ But as $a = c$ that means we have: :$\set {a, b} = \set {a, d}$ It follows from the lemma that: :$b = d$ {{qed}}
First a lemma: {{:Equality of Ordered Pairs/Lemma}}{{qed|lemma}} Let $\tuple {a, b} = \tuple {c, d}$. From the empty set formalization: :$\set {\set {\O, a}, \set {\set \O, b} } = \set {\set {\O, c}, \set {\set \O, d} }$ First we note the special case where $a = \set \O$ and $b = \O$. Then we have: {{begin-eqn}} {{eqn | l = \set {\set {\O, a}, \set {\set \O, b} }       | r = \set {\set {\O, \set \O}, \set {\set \O, \O} }       | c =  }} {{eqn | r = \set {\set {\O, \set \O}, \set {\O, \set \O} }       | c = Axiom of Extension: $\set {\O, \set \O } = \set {\set \O, \O}$ }} {{eqn | r = \set {\set {\O, \set \O} }       | c = Doubleton Class of Equal Sets is Singleton Class }} {{eqn | r = \set {\set {\O, c}, \set {\set \O, d} }       | c = by definition }} {{end-eqn}} Thus we have that: :$c = d = \set {\O, \set \O}$ and so: :$c = \set \O$ and $d = \O$ leading to the result that $a = c$ and $b = d$. {{qed|lemma}} Suppose otherwise, that either $a \ne \set \O$ or $b \ne \O$, or both. Let $x \in \set {\set {\O, a}, \set {\set \O, b} }$. Then either: :$x = \set {\O, a}$ or: :$x = \set {\set \O, b}$ First suppose $x = \set {\O, a}$. Then: :$x = \set {\O, c}$ or $x = \set {\set \O, d}$ {{AimForCont}} $x = \set {\set \O, d}$. That is: :$\set {\O, a} = \set {\set \O, d}$ which means that $a = \set \O$ and $d = \O$. But as $\O \ne \set \O$ it follows by Proof by Contradiction that $x \ne \set {\set \O, d}$. Thus $x = \set {\O, c}$. We have that: :$x = \set {\O, a}$ and $x = \set {\O, c}$ and so: :$\set {\O, a} = \set {\O, c}$ It follows from the lemma that: :$a = c$ Now suppose $x = \set {\set \O, b}$. Then: :$x = \set {\O, c} \lor \set {\set \O, d}$ {{AimForCont}} $x = \set {\O, c}$. That is: :$\set {\set \O, b} = \set {\O, c}$ which means that $b = \O$ and $c = \set \O$. But as $\O \ne \set \O$ it follows by Proof by Contradiction that $x \ne \set {\O, c}$. Thus $x = \set {\set \O, d}$. We have that: :$x = \set {\set \O, b}$ and $x = \set {\set \O, d}$ and so: :$\set {\set \O, b} = \set {\set \O, d}$ It follows from the lemma that: :$b = d$ {{qed}}
First a lemma: {{:Equality of Ordered Pairs/Lemma}}{{qed|lemma}} Let $\tuple {a, b} = \tuple {c, d}$. From the Wiener formalization: :$\set {\set {\O, \set a}, \set {\set b} } = \set {\set {\O, \set c}, \set {\set d} }$ Let $x \in \set {\set {\O, \set a}, \set {\set b} }$. Then either: :$x = \set {\O, \set a}$ or: :$x = \set {\set b}$ First suppose $x = \set {\O, \set a}$. Then: :$x = \set {\O, \set c}$ or $x = \set {\set d}$ {{AimForCont}} $x = \set {\set d}$. That is: :$\set {\O, \set a} = \set {\set d}$ Then: :$\set d = \set a$ and also: :$\set d = \O$ But that means: :$a \in \O$ which contradicts the definition of the empty set $\O$. It follows by Proof by Contradiction that $x \ne \set {\set d}$. Thus $x = \set {\O, \set c}$. We have that: :$x = \set {\O, \set a}$ and $x = \set {\O, \set c}$ and so: :$\set {\O, \set a} = \set {\O, \set c}$ It follows from the lemma that: :$\set a = \set c$ Hence from Singleton Equality: :$a = c$ Now suppose $x = \set {\set b}$. Then: :$x = \set {\O, \set c}$ or $x = \set {\set d}$ {{AimForCont}} $x = \set {\O, \set c}$. That is: :$\set {\set b} = \set {\O, \set c}$ which means that $\set b = \O$ and $\set b = \set c$. But as $\O \ne \set \O$ it follows by Proof by Contradiction that $x \ne \set {\O, \set c}$. Thus $x = \set {\set d}$. We have that: :$x = \set {\set b}$ and $x = \set {\set d}$ By Singleton Equality: :$\set b = \set d$ and again by Singleton Equality: :$b = d$ {{qed}}
We are given that $A$ is non-empty. Then $\exists x \in A$ where $x$ is a set. By definition of intersection of class, every element of $\displaystyle \bigcap A$ is an element of all elements of $A$. Thus: :$\displaystyle \bigcap A \subseteq x$ We are given that $A$ is a subclass of the basic universe $V$. Thus $x \in V$ by definition of basic universe. By the Axiom of Swelledness, $V$ is a swelled class. By definition of swelled class, every subclass of a set $x \in V$ is a set. It follows $\displaystyle \bigcap A$ is a set. {{Qed}}
By definition of $V$ we have that $\displaystyle \bigcap \O \subseteq V$. By definition of empty class, there exists no set $x \in V$ which is an elements of $\O$. Hence it is vacuously true that every element of $V$ is an element of every element of $\O$. Therefore every $x \in V$ is an element of $\displaystyle \bigcap \O$. Therefore $V \subseteq \displaystyle \bigcap \O$. It follows by definition of class equality that $\displaystyle \bigcap \O = V$. {{qed}}
By the axiom of unions, both $\displaystyle \bigcup A$ and $\displaystyle \bigcup B$ are sets. Let $x \in \displaystyle \bigcup A$. Then: :$\exists y \in A: x \in y$ But as $A \subseteq B$ it follows that $y \in B$. That is: :$\exists y \in B: x \in y$ That is: :$x \in \displaystyle \bigcup B$ Hence the result by definition of subset. {{Qed}}
First we consider the degenerate case where $A = B = \O$. By Intersection of Empty Class we have that: :$\displaystyle \bigcap \O = V$ Thus we have: :$\displaystyle \bigcap B = \displaystyle \bigcap A = V$ and neither $\displaystyle \bigcap B$ nor $\displaystyle \bigcap A$ are in fact sets. So while in this case $\displaystyle \bigcap B \subseteq \displaystyle \bigcap A$, $\displaystyle \bigcap B$ is a subclass of $\displaystyle \bigcap A$ and not actually a subset. {{qed|lemma}} Next we cover the case where $A = \O$ is the empty class. It is noted that Empty Class is Subclass of All Classes, and so $A \subseteq B$. By Intersection of Empty Class we have that: :$\displaystyle \bigcap A = V$ and so by definition of basic universe $\displaystyle \bigcap B \subseteq V$. By Intersection of Non-Empty Class is Set, $\displaystyle \bigcap B$ is a set. By definition of basic universe: :$\displaystyle \bigcap B \in V$ By the axiom of transitivity, $V$ is transitive. Hence as $\displaystyle \bigcap B \in V$ it follows that: :$\displaystyle \bigcap B \subseteq V$ and the result holds. {{qed|lemma}} By Intersection of Non-Empty Class is Set, both $\displaystyle \bigcap A$ and $\displaystyle \bigcap B$ are sets. Let $x \in \displaystyle \bigcap B$. Then: :$\forall y \in B: x \in y$ That is: :$y \in B \implies x \in y$ But as $A \subseteq B$, we have that: :$y \in A \implies y \in B$ and therefore: :$y \in A \implies x \in y$ That is: :$\forall y \in A: x \in y$ and so by definition of intersection: :$x \in \displaystyle \bigcap A$ Hence the result by definition of subset. {{Qed}}
$A$ be transitive. Let $x \in \displaystyle \bigcup A$. Then by definition: :$\exists y \in A: x \in y$ By definition of transitive class: :$x \in y \land y \in A \implies x \in A$ and so: :$x \in A$ Hence the result by definition of subclass.