Set Difference with Set Difference/Proof 2
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Theorem
- $S \setminus \paren {S \setminus T} = S \cap T = T \setminus \paren {T \setminus S}$
Proof
From the Axiom of Transitivity, all sets are classes.
The result then follows from Class Difference with Class Difference.
This needs considerable tedious hard slog to complete it. In particular: Find the result that demonstrate the set difference of two sets is also a set, intersection as well. Probably via subset of set is set or something. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
This page may be the result of a refactoring operation. As such, the following source works, along with any process flow, will need to be reviewed. When this has been completed, the citation of that source work (if it is appropriate that it stay on this page) is to be placed above this message, into the usual chronological ordering. If you have access to any of these works, then you are invited to review this list, and make any necessary corrections. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{SourceReview}} from the code. |
- 1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $2 \ \text{(a)}$
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 5$: Complements and Powers
- 1961: John G. Hocking and Gail S. Young: Topology ... (previous) ... (next): A Note on Set-Theoretic Concepts: $(5)$
- 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): Exercise $1.1: \ 6$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $1$: Sets and Logic: Exercise $5$
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 2$: Problem $1 \ \text{(iii)}$