Mixing
Proof verification for the equivalence of two sets.
$begingroup$
I have constructed two identical draft proofs for the following question using implications and words. Can you please verify whether they are logically correct. Should I have used De Morgan's Laws?
Exercise:
Suppose that $C,D$ are subsets of a set $X$. Prove that $$(Xsetminus C){,}cap{,}D =Dsetminus C.$$
Proof 1:
Suppose that $xin{(Xsetminus C){,}cap{,}D}$. Then $xin{(Xsetminus C)}$ and $xin{D}$. Then ($xin{X}$ and $xnotin{C}$) and $xin{D}$. Then ($xin{X}$ and $xin{D}$) and ($xin{D}$ and $xnotin {C}$). Then $xin(X cap D) cap (D setminus C)$. Thus, $xin(D setminus C)$. So, $(X setminus C) {,} cap D subseteq (D setminus C)$.
Conversely, suppose that $xin (Dsetminus C)$. Then $(xin{D}$ and $xnotin{C})$. Then $(xin X$ and $xin{D}$) and $xnotin{C}$. Then $(xin{X}$ and $xnotin{C})$ and $xin{D}$. Thus $xin(Xsetminus {C})cap{D}$. So, $(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}$.
Since $(X setminus C) {,} cap D subseteq (D setminus C)$ and $(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}$, we have that $(X setminus C) {,} cap D = (D setminus C)\$.
Proof 2:
Suppose that $xin{(Xsetminus C){,}cap{,}D}$. Then,
begin{align}
&implies xin{(Xsetminus C)}{,}{,}text{and}{,}D \
&implies(xin{X} {,}text{and} {,}xnotin{C}) {,}text{and}{,} xin{D} \
&implies (xin{X} {,}text{and}{,} xin{D}) {,}text{and} {,}(xin{D} {,}text{and}{,} xnotin {C})\
&implies xin(X cap D) cap (D setminus C)\
&implies xin(D setminus C).\ \
text{Thus}, (X setminus C) {,} cap D subseteq (D setminus C).\ \
end{align}
Conversely, suppose $xin (Dsetminus C)$. Then,
begin{align}
&implies (xin{D} {,}text{and}{,} xnotin{C}) \
&implies (xin X {,}text{and}{,} xin{D}){,}text{and}{,} xnotin{C} \
&implies (xin{X}{,}text{and}{,} xnotin{C}){,}text{and}{,} xin{D} \
&implies xin(Xsetminus {C})cap{D}. \ \
text{Thus,}{,}(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}.
end{align}
Since $(X setminus C) {,} cap D subseteq (D setminus C)$ and $(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}$, we have that $(X setminus C) {,} cap D = (D setminus C)$.
proof-verification elementary-set-theory logic
edited Jan 15 at 2:43
Andrés E. Caicedo
65.5k8158249
$endgroup$
add a comment |
$begingroup$
I have constructed two identical draft proofs for the following question using implications and words. Can you please verify whether they are logically correct. Should I have used De Morgan's Laws?
Exercise:
Suppose that $C,D$ are subsets of a set $X$. Prove that $$(Xsetminus C){,}cap{,}D =Dsetminus C.$$
Proof 1:
Suppose that $xin{(Xsetminus C){,}cap{,}D}$. Then $xin{(Xsetminus C)}$ and $xin{D}$. Then ($xin{X}$ and $xnotin{C}$) and $xin{D}$. Then ($xin{X}$ and $xin{D}$) and ($xin{D}$ and $xnotin {C}$). Then $xin(X cap D) cap (D setminus C)$. Thus, $xin(D setminus C)$. So, $(X setminus C) {,} cap D subseteq (D setminus C)$.
Conversely, suppose that $xin (Dsetminus C)$. Then $(xin{D}$ and $xnotin{C})$. Then $(xin X$ and $xin{D}$) and $xnotin{C}$. Then $(xin{X}$ and $xnotin{C})$ and $xin{D}$. Thus $xin(Xsetminus {C})cap{D}$. So, $(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}$.
Since $(X setminus C) {,} cap D subseteq (D setminus C)$ and $(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}$, we have that $(X setminus C) {,} cap D = (D setminus C)\$.
Proof 2:
Suppose that $xin{(Xsetminus C){,}cap{,}D}$. Then,
begin{align}
&implies xin{(Xsetminus C)}{,}{,}text{and}{,}D \
&implies(xin{X} {,}text{and} {,}xnotin{C}) {,}text{and}{,} xin{D} \
&implies (xin{X} {,}text{and}{,} xin{D}) {,}text{and} {,}(xin{D} {,}text{and}{,} xnotin {C})\
&implies xin(X cap D) cap (D setminus C)\
&implies xin(D setminus C).\ \
text{Thus}, (X setminus C) {,} cap D subseteq (D setminus C).\ \
end{align}
Conversely, suppose $xin (Dsetminus C)$. Then,
begin{align}
&implies (xin{D} {,}text{and}{,} xnotin{C}) \
&implies (xin X {,}text{and}{,} xin{D}){,}text{and}{,} xnotin{C} \
&implies (xin{X}{,}text{and}{,} xnotin{C}){,}text{and}{,} xin{D} \
&implies xin(Xsetminus {C})cap{D}. \ \
text{Thus,}{,}(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}.
end{align}
Since $(X setminus C) {,} cap D subseteq (D setminus C)$ and $(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}$, we have that $(X setminus C) {,} cap D = (D setminus C)$.
proof-verification elementary-set-theory logic
edited Jan 15 at 2:43
Andrés E. Caicedo
65.5k8158249
$endgroup$
add a comment |
$begingroup$
I have constructed two identical draft proofs for the following question using implications and words. Can you please verify whether they are logically correct. Should I have used De Morgan's Laws?
Exercise:
Suppose that $C,D$ are subsets of a set $X$. Prove that $$(Xsetminus C){,}cap{,}D =Dsetminus C.$$
Proof 1:
Suppose that $xin{(Xsetminus C){,}cap{,}D}$. Then $xin{(Xsetminus C)}$ and $xin{D}$. Then ($xin{X}$ and $xnotin{C}$) and $xin{D}$. Then ($xin{X}$ and $xin{D}$) and ($xin{D}$ and $xnotin {C}$). Then $xin(X cap D) cap (D setminus C)$. Thus, $xin(D setminus C)$. So, $(X setminus C) {,} cap D subseteq (D setminus C)$.
Conversely, suppose that $xin (Dsetminus C)$. Then $(xin{D}$ and $xnotin{C})$. Then $(xin X$ and $xin{D}$) and $xnotin{C}$. Then $(xin{X}$ and $xnotin{C})$ and $xin{D}$. Thus $xin(Xsetminus {C})cap{D}$. So, $(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}$.
Since $(X setminus C) {,} cap D subseteq (D setminus C)$ and $(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}$, we have that $(X setminus C) {,} cap D = (D setminus C)\$.
Proof 2:
Suppose that $xin{(Xsetminus C){,}cap{,}D}$. Then,
begin{align}
&implies xin{(Xsetminus C)}{,}{,}text{and}{,}D \
&implies(xin{X} {,}text{and} {,}xnotin{C}) {,}text{and}{,} xin{D} \
&implies (xin{X} {,}text{and}{,} xin{D}) {,}text{and} {,}(xin{D} {,}text{and}{,} xnotin {C})\
&implies xin(X cap D) cap (D setminus C)\
&implies xin(D setminus C).\ \
text{Thus}, (X setminus C) {,} cap D subseteq (D setminus C).\ \
end{align}
Conversely, suppose $xin (Dsetminus C)$. Then,
begin{align}
&implies (xin{D} {,}text{and}{,} xnotin{C}) \
&implies (xin X {,}text{and}{,} xin{D}){,}text{and}{,} xnotin{C} \
&implies (xin{X}{,}text{and}{,} xnotin{C}){,}text{and}{,} xin{D} \
&implies xin(Xsetminus {C})cap{D}. \ \
text{Thus,}{,}(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}.
end{align}
Since $(X setminus C) {,} cap D subseteq (D setminus C)$ and $(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}$, we have that $(X setminus C) {,} cap D = (D setminus C)$.
proof-verification elementary-set-theory logic
edited Jan 15 at 2:43
Andrés E. Caicedo
65.5k8158249
$endgroup$
I have constructed two identical draft proofs for the following question using implications and words. Can you please verify whether they are logically correct. Should I have used De Morgan's Laws?
Exercise:
Suppose that $C,D$ are subsets of a set $X$. Prove that $$(Xsetminus C){,}cap{,}D =Dsetminus C.$$
Proof 1:
Suppose that $xin{(Xsetminus C){,}cap{,}D}$. Then $xin{(Xsetminus C)}$ and $xin{D}$. Then ($xin{X}$ and $xnotin{C}$) and $xin{D}$. Then ($xin{X}$ and $xin{D}$) and ($xin{D}$ and $xnotin {C}$). Then $xin(X cap D) cap (D setminus C)$. Thus, $xin(D setminus C)$. So, $(X setminus C) {,} cap D subseteq (D setminus C)$.
Conversely, suppose that $xin (Dsetminus C)$. Then $(xin{D}$ and $xnotin{C})$. Then $(xin X$ and $xin{D}$) and $xnotin{C}$. Then $(xin{X}$ and $xnotin{C})$ and $xin{D}$. Thus $xin(Xsetminus {C})cap{D}$. So, $(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}$.
Since $(X setminus C) {,} cap D subseteq (D setminus C)$ and $(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}$, we have that $(X setminus C) {,} cap D = (D setminus C)\$.
Proof 2:
Suppose that $xin{(Xsetminus C){,}cap{,}D}$. Then,
begin{align}
&implies xin{(Xsetminus C)}{,}{,}text{and}{,}D \
&implies(xin{X} {,}text{and} {,}xnotin{C}) {,}text{and}{,} xin{D} \
&implies (xin{X} {,}text{and}{,} xin{D}) {,}text{and} {,}(xin{D} {,}text{and}{,} xnotin {C})\
&implies xin(X cap D) cap (D setminus C)\
&implies xin(D setminus C).\ \
text{Thus}, (X setminus C) {,} cap D subseteq (D setminus C).\ \
end{align}
Conversely, suppose $xin (Dsetminus C)$. Then,
begin{align}
&implies (xin{D} {,}text{and}{,} xnotin{C}) \
&implies (xin X {,}text{and}{,} xin{D}){,}text{and}{,} xnotin{C} \
&implies (xin{X}{,}text{and}{,} xnotin{C}){,}text{and}{,} xin{D} \
&implies xin(Xsetminus {C})cap{D}. \ \
text{Thus,}{,}(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}.
end{align}
Since $(X setminus C) {,} cap D subseteq (D setminus C)$ and $(Dsetminus{C}) subseteq{(Xsetminus{C}})cap{D}$, we have that $(X setminus C) {,} cap D = (D setminus C)$.
proof-verification elementary-set-theory logic
proof-verification elementary-set-theory logic
edited Jan 15 at 2:43
Andrés E. Caicedo
65.5k8158249
edited Jan 15 at 2:43
Andrés E. Caicedo
65.5k8158249
edited Jan 15 at 2:43
Andrés E. Caicedo
65.5k8158249
edited Jan 15 at 2:43
Andrés E. Caicedo
65.5k8158249
edited Jan 15 at 2:43
Andrés E. Caicedo
65.5k8158249
65.5k8158249
asked Jan 15 at 2:06
user503154
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2 Answers
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$begingroup$
Your logic all seems to follow fine. My main nitpick would be to more clearly indicate when the assumption $C,D subseteq X$ is used.
With respect to whether you should have used De Morgan's law, I don't think they would have made anything easier. I'm not even sure if they would have any use here. In any event your method of proof is how I would've done things at any rate.
answered Jan 15 at 2:15
Eevee TrainerEevee Trainer
6,0931936
$endgroup$
2
$begingroup$
Thanks. Can you please let me know where you would indicate in the proof what you have suggested. :)
$endgroup$
– user503154
Jan 15 at 2:25
1
$begingroup$
The main point that it came to mind was when you went from $$(xin{D} {,}text{and}{,} xnotin{C})$$ to $$(xin X {,}text{and}{,} xin{D}){,}text{and}{,} xnotin{C}$$ It took me a bit to realize that this followed because $D subseteq X$. I didn't really have any other problems following the proof other than that, so that'd probably be the main point to apply it.
$endgroup$
– Eevee Trainer
Jan 15 at 2:31
add a comment |
$begingroup$
Perhaps more briefly:
For any sets $A,B$ we define $Acap B$ by $forall x,(;xin Acap Biff (xin Aland xin B);).$ And we define $Asetminus B$ by $;forall x ,(;xin Asetminus B iff (xin Aland x not in B);).$
And we have $Dsubset Xiff forall x;(xin D iff (xin Dland xin X))$ .
So if $Dsubset X$ then for all $x$ we have $$[xin Dsetminus C]iff$$ $$iff [xin Dland xnot in C] iff$$ $$iff [(xin Dland xin X)land xnot in C] iff$$ $$iff [xin D land (xin Xland x not in C)]iff$$ $$iff [xin Dland xin Xsetminus C] iff$$ $$iff [ xin Dcap (Xsetminus C)].$$ And it doesn't matter whether or not $Csubset X.$
answered Jan 15 at 7:29
DanielWainfleetDanielWainfleet
35.1k31648
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Your logic all seems to follow fine. My main nitpick would be to more clearly indicate when the assumption $C,D subseteq X$ is used.
With respect to whether you should have used De Morgan's law, I don't think they would have made anything easier. I'm not even sure if they would have any use here. In any event your method of proof is how I would've done things at any rate.
answered Jan 15 at 2:15
Eevee TrainerEevee Trainer
6,0931936
$endgroup$
2
$begingroup$
Thanks. Can you please let me know where you would indicate in the proof what you have suggested. :)
$endgroup$
– user503154
Jan 15 at 2:25
1
$begingroup$
The main point that it came to mind was when you went from $$(xin{D} {,}text{and}{,} xnotin{C})$$ to $$(xin X {,}text{and}{,} xin{D}){,}text{and}{,} xnotin{C}$$ It took me a bit to realize that this followed because $D subseteq X$. I didn't really have any other problems following the proof other than that, so that'd probably be the main point to apply it.
$endgroup$
– Eevee Trainer
Jan 15 at 2:31
add a comment |
$begingroup$
Your logic all seems to follow fine. My main nitpick would be to more clearly indicate when the assumption $C,D subseteq X$ is used.
With respect to whether you should have used De Morgan's law, I don't think they would have made anything easier. I'm not even sure if they would have any use here. In any event your method of proof is how I would've done things at any rate.
answered Jan 15 at 2:15
Eevee TrainerEevee Trainer
6,0931936
$endgroup$
2
$begingroup$
Thanks. Can you please let me know where you would indicate in the proof what you have suggested. :)
$endgroup$
– user503154
Jan 15 at 2:25
1
$begingroup$
The main point that it came to mind was when you went from $$(xin{D} {,}text{and}{,} xnotin{C})$$ to $$(xin X {,}text{and}{,} xin{D}){,}text{and}{,} xnotin{C}$$ It took me a bit to realize that this followed because $D subseteq X$. I didn't really have any other problems following the proof other than that, so that'd probably be the main point to apply it.
$endgroup$
– Eevee Trainer
Jan 15 at 2:31
add a comment |
$begingroup$
Your logic all seems to follow fine. My main nitpick would be to more clearly indicate when the assumption $C,D subseteq X$ is used.
With respect to whether you should have used De Morgan's law, I don't think they would have made anything easier. I'm not even sure if they would have any use here. In any event your method of proof is how I would've done things at any rate.
answered Jan 15 at 2:15
Eevee TrainerEevee Trainer
6,0931936
$endgroup$
Your logic all seems to follow fine. My main nitpick would be to more clearly indicate when the assumption $C,D subseteq X$ is used.
With respect to whether you should have used De Morgan's law, I don't think they would have made anything easier. I'm not even sure if they would have any use here. In any event your method of proof is how I would've done things at any rate.
answered Jan 15 at 2:15
Eevee TrainerEevee Trainer
6,0931936
answered Jan 15 at 2:15
Eevee TrainerEevee Trainer
6,0931936
answered Jan 15 at 2:15
Eevee TrainerEevee Trainer
6,0931936
answered Jan 15 at 2:15
Eevee TrainerEevee Trainer
6,0931936
6,0931936
2
$begingroup$
Thanks. Can you please let me know where you would indicate in the proof what you have suggested. :)
$endgroup$
– user503154
Jan 15 at 2:25
1
$begingroup$
The main point that it came to mind was when you went from $$(xin{D} {,}text{and}{,} xnotin{C})$$ to $$(xin X {,}text{and}{,} xin{D}){,}text{and}{,} xnotin{C}$$ It took me a bit to realize that this followed because $D subseteq X$. I didn't really have any other problems following the proof other than that, so that'd probably be the main point to apply it.
$endgroup$
– Eevee Trainer
Jan 15 at 2:31
add a comment |
2
$begingroup$
Thanks. Can you please let me know where you would indicate in the proof what you have suggested. :)
$endgroup$
– user503154
Jan 15 at 2:25
1
$begingroup$
The main point that it came to mind was when you went from $$(xin{D} {,}text{and}{,} xnotin{C})$$ to $$(xin X {,}text{and}{,} xin{D}){,}text{and}{,} xnotin{C}$$ It took me a bit to realize that this followed because $D subseteq X$. I didn't really have any other problems following the proof other than that, so that'd probably be the main point to apply it.
$endgroup$
– Eevee Trainer
Jan 15 at 2:31
2
2
$begingroup$
Thanks. Can you please let me know where you would indicate in the proof what you have suggested. :)
$endgroup$
– user503154
Jan 15 at 2:25
$begingroup$
Thanks. Can you please let me know where you would indicate in the proof what you have suggested. :)
$endgroup$
– user503154
Jan 15 at 2:25
1
1
$begingroup$
The main point that it came to mind was when you went from $$(xin{D} {,}text{and}{,} xnotin{C})$$ to $$(xin X {,}text{and}{,} xin{D}){,}text{and}{,} xnotin{C}$$ It took me a bit to realize that this followed because $D subseteq X$. I didn't really have any other problems following the proof other than that, so that'd probably be the main point to apply it.
$endgroup$
– Eevee Trainer
Jan 15 at 2:31
$begingroup$
The main point that it came to mind was when you went from $$(xin{D} {,}text{and}{,} xnotin{C})$$ to $$(xin X {,}text{and}{,} xin{D}){,}text{and}{,} xnotin{C}$$ It took me a bit to realize that this followed because $D subseteq X$. I didn't really have any other problems following the proof other than that, so that'd probably be the main point to apply it.
$endgroup$
– Eevee Trainer
Jan 15 at 2:31
add a comment |
$begingroup$
Perhaps more briefly:
For any sets $A,B$ we define $Acap B$ by $forall x,(;xin Acap Biff (xin Aland xin B);).$ And we define $Asetminus B$ by $;forall x ,(;xin Asetminus B iff (xin Aland x not in B);).$
And we have $Dsubset Xiff forall x;(xin D iff (xin Dland xin X))$ .
So if $Dsubset X$ then for all $x$ we have $$[xin Dsetminus C]iff$$ $$iff [xin Dland xnot in C] iff$$ $$iff [(xin Dland xin X)land xnot in C] iff$$ $$iff [xin D land (xin Xland x not in C)]iff$$ $$iff [xin Dland xin Xsetminus C] iff$$ $$iff [ xin Dcap (Xsetminus C)].$$ And it doesn't matter whether or not $Csubset X.$
answered Jan 15 at 7:29
DanielWainfleetDanielWainfleet
35.1k31648
$endgroup$
add a comment |
$begingroup$
Perhaps more briefly:
For any sets $A,B$ we define $Acap B$ by $forall x,(;xin Acap Biff (xin Aland xin B);).$ And we define $Asetminus B$ by $;forall x ,(;xin Asetminus B iff (xin Aland x not in B);).$
And we have $Dsubset Xiff forall x;(xin D iff (xin Dland xin X))$ .
So if $Dsubset X$ then for all $x$ we have $$[xin Dsetminus C]iff$$ $$iff [xin Dland xnot in C] iff$$ $$iff [(xin Dland xin X)land xnot in C] iff$$ $$iff [xin D land (xin Xland x not in C)]iff$$ $$iff [xin Dland xin Xsetminus C] iff$$ $$iff [ xin Dcap (Xsetminus C)].$$ And it doesn't matter whether or not $Csubset X.$
answered Jan 15 at 7:29
DanielWainfleetDanielWainfleet
35.1k31648
$endgroup$
add a comment |
$begingroup$
Perhaps more briefly:
For any sets $A,B$ we define $Acap B$ by $forall x,(;xin Acap Biff (xin Aland xin B);).$ And we define $Asetminus B$ by $;forall x ,(;xin Asetminus B iff (xin Aland x not in B);).$
And we have $Dsubset Xiff forall x;(xin D iff (xin Dland xin X))$ .
So if $Dsubset X$ then for all $x$ we have $$[xin Dsetminus C]iff$$ $$iff [xin Dland xnot in C] iff$$ $$iff [(xin Dland xin X)land xnot in C] iff$$ $$iff [xin D land (xin Xland x not in C)]iff$$ $$iff [xin Dland xin Xsetminus C] iff$$ $$iff [ xin Dcap (Xsetminus C)].$$ And it doesn't matter whether or not $Csubset X.$
answered Jan 15 at 7:29
DanielWainfleetDanielWainfleet
35.1k31648
$endgroup$
Perhaps more briefly:
For any sets $A,B$ we define $Acap B$ by $forall x,(;xin Acap Biff (xin Aland xin B);).$ And we define $Asetminus B$ by $;forall x ,(;xin Asetminus B iff (xin Aland x not in B);).$
And we have $Dsubset Xiff forall x;(xin D iff (xin Dland xin X))$ .
So if $Dsubset X$ then for all $x$ we have $$[xin Dsetminus C]iff$$ $$iff [xin Dland xnot in C] iff$$ $$iff [(xin Dland xin X)land xnot in C] iff$$ $$iff [xin D land (xin Xland x not in C)]iff$$ $$iff [xin Dland xin Xsetminus C] iff$$ $$iff [ xin Dcap (Xsetminus C)].$$ And it doesn't matter whether or not $Csubset X.$
answered Jan 15 at 7:29
DanielWainfleetDanielWainfleet
35.1k31648
answered Jan 15 at 7:29
DanielWainfleetDanielWainfleet
35.1k31648
answered Jan 15 at 7:29
DanielWainfleetDanielWainfleet
35.1k31648
answered Jan 15 at 7:29
DanielWainfleetDanielWainfleet
35.1k31648
35.1k31648
add a comment |
add a comment |
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