Prove, that for all sets A, B, C and D, in which $A subseteq B$ and $C subseteq D$, given formula: $A...
Knowing
$$A subseteq B Leftrightarrow (xin A Longrightarrow xin B)$$
$$C subseteq D Leftrightarrow (xin C Longrightarrow xin D)\$$
I tried to prove, that for any $yin A$ and $zin D$, given formula
$$A setminus D subseteq Bsetminus C$$
is true, but I got stuck almost at the beginning, when I tried to show $$(y,z)in Asetminus Dequiv {yin A}wedge {znotin D}$$ and I am not really sure what should I do now. Although I know, that if $xnotin D$, then $xnotin C$, but it really leads me to nowhere. I would like to see a formal proof of this. Thanks in advance.
elementary-set-theory
asked Nov 25 at 14:41
whiskeyo
435
add a comment |
Knowing
$$A subseteq B Leftrightarrow (xin A Longrightarrow xin B)$$
$$C subseteq D Leftrightarrow (xin C Longrightarrow xin D)\$$
I tried to prove, that for any $yin A$ and $zin D$, given formula
$$A setminus D subseteq Bsetminus C$$
is true, but I got stuck almost at the beginning, when I tried to show $$(y,z)in Asetminus Dequiv {yin A}wedge {znotin D}$$ and I am not really sure what should I do now. Although I know, that if $xnotin D$, then $xnotin C$, but it really leads me to nowhere. I would like to see a formal proof of this. Thanks in advance.
elementary-set-theory
asked Nov 25 at 14:41
whiskeyo
435
1
It's supposed to be $yin Asetminus Diff {yin A}wedge {ynotin D}$. No $z$ anywhere.
– Arthur
Nov 25 at 14:47
Thank you, I was trying to prove it the same way as I did for $Acup C subseteq Bcup D$, that's why I used two elements.
– whiskeyo
Nov 25 at 15:01
add a comment |
Knowing
$$A subseteq B Leftrightarrow (xin A Longrightarrow xin B)$$
$$C subseteq D Leftrightarrow (xin C Longrightarrow xin D)\$$
I tried to prove, that for any $yin A$ and $zin D$, given formula
$$A setminus D subseteq Bsetminus C$$
is true, but I got stuck almost at the beginning, when I tried to show $$(y,z)in Asetminus Dequiv {yin A}wedge {znotin D}$$ and I am not really sure what should I do now. Although I know, that if $xnotin D$, then $xnotin C$, but it really leads me to nowhere. I would like to see a formal proof of this. Thanks in advance.
elementary-set-theory
asked Nov 25 at 14:41
whiskeyo
435
Knowing
$$A subseteq B Leftrightarrow (xin A Longrightarrow xin B)$$
$$C subseteq D Leftrightarrow (xin C Longrightarrow xin D)\$$
I tried to prove, that for any $yin A$ and $zin D$, given formula
$$A setminus D subseteq Bsetminus C$$
is true, but I got stuck almost at the beginning, when I tried to show $$(y,z)in Asetminus Dequiv {yin A}wedge {znotin D}$$ and I am not really sure what should I do now. Although I know, that if $xnotin D$, then $xnotin C$, but it really leads me to nowhere. I would like to see a formal proof of this. Thanks in advance.
elementary-set-theory
elementary-set-theory
asked Nov 25 at 14:41
whiskeyo
435
asked Nov 25 at 14:41
whiskeyo
435
asked Nov 25 at 14:41
whiskeyo
435
asked Nov 25 at 14:41
whiskeyo
435
asked Nov 25 at 14:41
whiskeyo
435
435
1
It's supposed to be $yin Asetminus Diff {yin A}wedge {ynotin D}$. No $z$ anywhere.
– Arthur
Nov 25 at 14:47
Thank you, I was trying to prove it the same way as I did for $Acup C subseteq Bcup D$, that's why I used two elements.
– whiskeyo
Nov 25 at 15:01
add a comment |
1
It's supposed to be $yin Asetminus Diff {yin A}wedge {ynotin D}$. No $z$ anywhere.
– Arthur
Nov 25 at 14:47
Thank you, I was trying to prove it the same way as I did for $Acup C subseteq Bcup D$, that's why I used two elements.
– whiskeyo
Nov 25 at 15:01
1
1
It's supposed to be $yin Asetminus Diff {yin A}wedge {ynotin D}$. No $z$ anywhere.
– Arthur
Nov 25 at 14:47
It's supposed to be $yin Asetminus Diff {yin A}wedge {ynotin D}$. No $z$ anywhere.
– Arthur
Nov 25 at 14:47
Thank you, I was trying to prove it the same way as I did for $Acup C subseteq Bcup D$, that's why I used two elements.
– whiskeyo
Nov 25 at 15:01
Thank you, I was trying to prove it the same way as I did for $Acup C subseteq Bcup D$, that's why I used two elements.
– whiskeyo
Nov 25 at 15:01
add a comment |
2 Answers
2
active
oldest
votes
The problem is in your assumption: $$(y,z) in A setminus D$$
When showing membership for a non-relational set we suppose a single element in the set, not an ordered pair. This is how it should go:
$$xin (Asetminus D)implies xin A land x notin D$$
Notice, $Csubseteq D implies (forall x)[(x in C)rightarrow (x in D)]$, consequently, by modus tollens $xnotin D implies xnotin C$
Since $A subseteq B land C subseteq D$, then:
$$(xin A implies xin B) land (xnotin D implies xnotin C)$$
Therefore, $$x in A setminus D implies xin B setminus C$$
$$Asetminus D subseteq Bsetminus C$$
answered Nov 25 at 14:56
Bertrand Wittgenstein's Ghost
352114
add a comment |
$C subseteq D Leftrightarrow D^{complement} subseteq C^{complement}$
So $A setminus D = A cap D^{complement} subseteq B cap C^{complement} = B setminus C$ as $A subseteq B$ as well, and if we replace two sets in the intersection by posibly larger sets, we get a possibly larger set.
answered Nov 25 at 15:12
Henno Brandsma
104k346113
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
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votes
The problem is in your assumption: $$(y,z) in A setminus D$$
When showing membership for a non-relational set we suppose a single element in the set, not an ordered pair. This is how it should go:
$$xin (Asetminus D)implies xin A land x notin D$$
Notice, $Csubseteq D implies (forall x)[(x in C)rightarrow (x in D)]$, consequently, by modus tollens $xnotin D implies xnotin C$
Since $A subseteq B land C subseteq D$, then:
$$(xin A implies xin B) land (xnotin D implies xnotin C)$$
Therefore, $$x in A setminus D implies xin B setminus C$$
$$Asetminus D subseteq Bsetminus C$$
answered Nov 25 at 14:56
Bertrand Wittgenstein's Ghost
352114
add a comment |
The problem is in your assumption: $$(y,z) in A setminus D$$
When showing membership for a non-relational set we suppose a single element in the set, not an ordered pair. This is how it should go:
$$xin (Asetminus D)implies xin A land x notin D$$
Notice, $Csubseteq D implies (forall x)[(x in C)rightarrow (x in D)]$, consequently, by modus tollens $xnotin D implies xnotin C$
Since $A subseteq B land C subseteq D$, then:
$$(xin A implies xin B) land (xnotin D implies xnotin C)$$
Therefore, $$x in A setminus D implies xin B setminus C$$
$$Asetminus D subseteq Bsetminus C$$
answered Nov 25 at 14:56
Bertrand Wittgenstein's Ghost
352114
add a comment |
The problem is in your assumption: $$(y,z) in A setminus D$$
When showing membership for a non-relational set we suppose a single element in the set, not an ordered pair. This is how it should go:
$$xin (Asetminus D)implies xin A land x notin D$$
Notice, $Csubseteq D implies (forall x)[(x in C)rightarrow (x in D)]$, consequently, by modus tollens $xnotin D implies xnotin C$
Since $A subseteq B land C subseteq D$, then:
$$(xin A implies xin B) land (xnotin D implies xnotin C)$$
Therefore, $$x in A setminus D implies xin B setminus C$$
$$Asetminus D subseteq Bsetminus C$$
answered Nov 25 at 14:56
Bertrand Wittgenstein's Ghost
352114
The problem is in your assumption: $$(y,z) in A setminus D$$
When showing membership for a non-relational set we suppose a single element in the set, not an ordered pair. This is how it should go:
$$xin (Asetminus D)implies xin A land x notin D$$
Notice, $Csubseteq D implies (forall x)[(x in C)rightarrow (x in D)]$, consequently, by modus tollens $xnotin D implies xnotin C$
Since $A subseteq B land C subseteq D$, then:
$$(xin A implies xin B) land (xnotin D implies xnotin C)$$
Therefore, $$x in A setminus D implies xin B setminus C$$
$$Asetminus D subseteq Bsetminus C$$
answered Nov 25 at 14:56
Bertrand Wittgenstein's Ghost
352114
edited Nov 25 at 15:01
answered Nov 25 at 14:56
Bertrand Wittgenstein's Ghost
352114
answered Nov 25 at 14:56
Bertrand Wittgenstein's Ghost
352114
answered Nov 25 at 14:56
Bertrand Wittgenstein's Ghost
352114
352114
add a comment |
add a comment |
$C subseteq D Leftrightarrow D^{complement} subseteq C^{complement}$
So $A setminus D = A cap D^{complement} subseteq B cap C^{complement} = B setminus C$ as $A subseteq B$ as well, and if we replace two sets in the intersection by posibly larger sets, we get a possibly larger set.
answered Nov 25 at 15:12
Henno Brandsma
104k346113
add a comment |
$C subseteq D Leftrightarrow D^{complement} subseteq C^{complement}$
So $A setminus D = A cap D^{complement} subseteq B cap C^{complement} = B setminus C$ as $A subseteq B$ as well, and if we replace two sets in the intersection by posibly larger sets, we get a possibly larger set.
answered Nov 25 at 15:12
Henno Brandsma
104k346113
add a comment |
$C subseteq D Leftrightarrow D^{complement} subseteq C^{complement}$
So $A setminus D = A cap D^{complement} subseteq B cap C^{complement} = B setminus C$ as $A subseteq B$ as well, and if we replace two sets in the intersection by posibly larger sets, we get a possibly larger set.
answered Nov 25 at 15:12
Henno Brandsma
104k346113
$C subseteq D Leftrightarrow D^{complement} subseteq C^{complement}$
So $A setminus D = A cap D^{complement} subseteq B cap C^{complement} = B setminus C$ as $A subseteq B$ as well, and if we replace two sets in the intersection by posibly larger sets, we get a possibly larger set.
answered Nov 25 at 15:12
Henno Brandsma
104k346113
answered Nov 25 at 15:12
Henno Brandsma
104k346113
answered Nov 25 at 15:12
Henno Brandsma
104k346113
answered Nov 25 at 15:12
Henno Brandsma
104k346113
104k346113
add a comment |
add a comment |
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1
It's supposed to be $yin Asetminus Diff {yin A}wedge {ynotin D}$. No $z$ anywhere.
– Arthur
Nov 25 at 14:47
Thank you, I was trying to prove it the same way as I did for $Acup C subseteq Bcup D$, that's why I used two elements.
– whiskeyo
Nov 25 at 15:01