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21 May 2022
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N 14:12 | Subclass of Well-Ordered Class is Well-Ordered 8 changes history +1,835 [Prime.mover (8×)] | |||
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14:12 (cur | prev) −4 Prime.mover talk contribs | ||||
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13:47 (cur | prev) +100 Prime.mover talk contribs | ||||
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13:17 (cur | prev) +192 Prime.mover talk contribs | ||||
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13:14 (cur | prev) −47 Prime.mover talk contribs | |||
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13:10 (cur | prev) −23 Prime.mover talk contribs Tag: Manual revert | |||
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11:52 (cur | prev) +23 Prime.mover talk contribs Tag: Reverted | ||||
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09:30 (cur | prev) −10 Prime.mover talk contribs | |||
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09:29 (cur | prev) +1,604 Prime.mover talk contribs Created page with "== Theorem == Let $V$ be a basic universe. Let $\RR \subseteq V \times V$ be a relation. Let $A$ be a subclass of $V$ which is well-ordered under $\RR$. Let $B$ be a non-empty class subclass of $A$. Then $B$ is also well-ordered under $\RR$. == Proof..." | |||
| N 09:10 | Subclass of Subclass is Subclass diffhist +582 Prime.mover talk contribs Created page with "== Theorem == Let $A$, $B$ and $C$ be classes. Let $A$ be a subclass of $B$. Let $B$ be a subclass of $C$. Then $A$ is a subclass of $C$. == Proof == Let $x \in A$ be arbitrary. It follows by definition of subclass that $x \in B$. It further follows by definition of subclass that $x \in C$. So we have that $..." | ||||