Proof verification: subsets with symmetric difference

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Problem



Prove that $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$



where $A mathbin Delta B = (A-B) cup (B-A)$



Proof



Let $x$ be arbitrary and suppose $x in A mathbin Delta C$. For $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$ to be true, either $x in (A mathbin Delta B)$ or $x in (B mathbin Delta C)$. If $x in (A mathbin Delta B)$, obviously $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$ is true since $x$ was arbitrary. Now suppose $x notin (A mathbin Delta B)$



Case 1:



Suppose $x in A land x notin C$. Then $x in B$ for the following reason.



$x notin A mathbin Delta B$ means $(x notin A lor x in B)$ and $(x notin B lor xin A)$ Since $x in A$, for $(x notin A lor x in B)$ to be true, $x in B$



Since $x in B$ and $x notin C$, $x in B mathbin Delta C$



Case 2:



Suppose $x in C land x notin A$. By below $x notin B$.



Since $x notin A$, for $(x notin B lor xin A)$ to be true, $x notin B$



Since $x in C$ and $x notin B$, $x in B Delta C$



Since $x$ was arbitrary, $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$



I guess my biggest worry here is that my case may not be exhaustive.



Thanks for your time.







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  • I read it over and it looks good. Your two cases are indeed exhaustive: Since $x$ is assumed to be in $A triangle C$, this means $x in (A-C) cup (C-A)$. As you noted, $x in A-C$ means $x in A$ and $x notin C$, and likewise for $x in C-A$.
    – Lucas
    Jul 21 at 22:37










  • Thank you @Lucas
    – Phil
    Jul 22 at 2:52














up vote
0
down vote

favorite












Problem



Prove that $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$



where $A mathbin Delta B = (A-B) cup (B-A)$



Proof



Let $x$ be arbitrary and suppose $x in A mathbin Delta C$. For $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$ to be true, either $x in (A mathbin Delta B)$ or $x in (B mathbin Delta C)$. If $x in (A mathbin Delta B)$, obviously $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$ is true since $x$ was arbitrary. Now suppose $x notin (A mathbin Delta B)$



Case 1:



Suppose $x in A land x notin C$. Then $x in B$ for the following reason.



$x notin A mathbin Delta B$ means $(x notin A lor x in B)$ and $(x notin B lor xin A)$ Since $x in A$, for $(x notin A lor x in B)$ to be true, $x in B$



Since $x in B$ and $x notin C$, $x in B mathbin Delta C$



Case 2:



Suppose $x in C land x notin A$. By below $x notin B$.



Since $x notin A$, for $(x notin B lor xin A)$ to be true, $x notin B$



Since $x in C$ and $x notin B$, $x in B Delta C$



Since $x$ was arbitrary, $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$



I guess my biggest worry here is that my case may not be exhaustive.



Thanks for your time.







share|cite|improve this question





















  • I read it over and it looks good. Your two cases are indeed exhaustive: Since $x$ is assumed to be in $A triangle C$, this means $x in (A-C) cup (C-A)$. As you noted, $x in A-C$ means $x in A$ and $x notin C$, and likewise for $x in C-A$.
    – Lucas
    Jul 21 at 22:37










  • Thank you @Lucas
    – Phil
    Jul 22 at 2:52












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Problem



Prove that $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$



where $A mathbin Delta B = (A-B) cup (B-A)$



Proof



Let $x$ be arbitrary and suppose $x in A mathbin Delta C$. For $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$ to be true, either $x in (A mathbin Delta B)$ or $x in (B mathbin Delta C)$. If $x in (A mathbin Delta B)$, obviously $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$ is true since $x$ was arbitrary. Now suppose $x notin (A mathbin Delta B)$



Case 1:



Suppose $x in A land x notin C$. Then $x in B$ for the following reason.



$x notin A mathbin Delta B$ means $(x notin A lor x in B)$ and $(x notin B lor xin A)$ Since $x in A$, for $(x notin A lor x in B)$ to be true, $x in B$



Since $x in B$ and $x notin C$, $x in B mathbin Delta C$



Case 2:



Suppose $x in C land x notin A$. By below $x notin B$.



Since $x notin A$, for $(x notin B lor xin A)$ to be true, $x notin B$



Since $x in C$ and $x notin B$, $x in B Delta C$



Since $x$ was arbitrary, $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$



I guess my biggest worry here is that my case may not be exhaustive.



Thanks for your time.







share|cite|improve this question













Problem



Prove that $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$



where $A mathbin Delta B = (A-B) cup (B-A)$



Proof



Let $x$ be arbitrary and suppose $x in A mathbin Delta C$. For $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$ to be true, either $x in (A mathbin Delta B)$ or $x in (B mathbin Delta C)$. If $x in (A mathbin Delta B)$, obviously $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$ is true since $x$ was arbitrary. Now suppose $x notin (A mathbin Delta B)$



Case 1:



Suppose $x in A land x notin C$. Then $x in B$ for the following reason.



$x notin A mathbin Delta B$ means $(x notin A lor x in B)$ and $(x notin B lor xin A)$ Since $x in A$, for $(x notin A lor x in B)$ to be true, $x in B$



Since $x in B$ and $x notin C$, $x in B mathbin Delta C$



Case 2:



Suppose $x in C land x notin A$. By below $x notin B$.



Since $x notin A$, for $(x notin B lor xin A)$ to be true, $x notin B$



Since $x in C$ and $x notin B$, $x in B Delta C$



Since $x$ was arbitrary, $A mathbin Delta C subseteq (A mathbin Delta B) cup (B mathbin Delta C)$



I guess my biggest worry here is that my case may not be exhaustive.



Thanks for your time.









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share|cite|improve this question




share|cite|improve this question








edited Jul 21 at 18:38









Michael Hardy

204k23186462




204k23186462









asked Jul 21 at 18:15









Phil

2716




2716











  • I read it over and it looks good. Your two cases are indeed exhaustive: Since $x$ is assumed to be in $A triangle C$, this means $x in (A-C) cup (C-A)$. As you noted, $x in A-C$ means $x in A$ and $x notin C$, and likewise for $x in C-A$.
    – Lucas
    Jul 21 at 22:37










  • Thank you @Lucas
    – Phil
    Jul 22 at 2:52
















  • I read it over and it looks good. Your two cases are indeed exhaustive: Since $x$ is assumed to be in $A triangle C$, this means $x in (A-C) cup (C-A)$. As you noted, $x in A-C$ means $x in A$ and $x notin C$, and likewise for $x in C-A$.
    – Lucas
    Jul 21 at 22:37










  • Thank you @Lucas
    – Phil
    Jul 22 at 2:52















I read it over and it looks good. Your two cases are indeed exhaustive: Since $x$ is assumed to be in $A triangle C$, this means $x in (A-C) cup (C-A)$. As you noted, $x in A-C$ means $x in A$ and $x notin C$, and likewise for $x in C-A$.
– Lucas
Jul 21 at 22:37




I read it over and it looks good. Your two cases are indeed exhaustive: Since $x$ is assumed to be in $A triangle C$, this means $x in (A-C) cup (C-A)$. As you noted, $x in A-C$ means $x in A$ and $x notin C$, and likewise for $x in C-A$.
– Lucas
Jul 21 at 22:37












Thank you @Lucas
– Phil
Jul 22 at 2:52




Thank you @Lucas
– Phil
Jul 22 at 2:52















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