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.
proof-verification proof-writing
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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.
proof-verification proof-writing
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
add a comment |Â
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.
proof-verification proof-writing
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.
proof-verification proof-writing
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
add a comment |Â
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
add a comment |Â
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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