Proving a set statement











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A, B, C are subsets of a set U:



A ⊆ B → A ∩ B $nsubseteq$ C (1)



A ⊆ B ∨ A ⊆ C (2)



A ∩ C ⊆ B (3)



I have to prove that this is valid:



A ∩ B $nsubseteq$ C (4)



It is recommended to use this in our proof:



X ⊆ Y ↔ X ∩ Y = X (5)



Should this be solved by mathematical induction or somehow else? I don't know what to do about that. Sorry for bad English. That is not my first language.










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  • It would be much better if you showed how much work you have done, and where exactly are you stuck. Which statements are premises, what do you want proven... so forth.
    – Bertrand Wittgenstein's Ghost
    Nov 29 at 8:40

















up vote
-1
down vote

favorite












A, B, C are subsets of a set U:



A ⊆ B → A ∩ B $nsubseteq$ C (1)



A ⊆ B ∨ A ⊆ C (2)



A ∩ C ⊆ B (3)



I have to prove that this is valid:



A ∩ B $nsubseteq$ C (4)



It is recommended to use this in our proof:



X ⊆ Y ↔ X ∩ Y = X (5)



Should this be solved by mathematical induction or somehow else? I don't know what to do about that. Sorry for bad English. That is not my first language.










share|cite|improve this question






















  • It would be much better if you showed how much work you have done, and where exactly are you stuck. Which statements are premises, what do you want proven... so forth.
    – Bertrand Wittgenstein's Ghost
    Nov 29 at 8:40















up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











A, B, C are subsets of a set U:



A ⊆ B → A ∩ B $nsubseteq$ C (1)



A ⊆ B ∨ A ⊆ C (2)



A ∩ C ⊆ B (3)



I have to prove that this is valid:



A ∩ B $nsubseteq$ C (4)



It is recommended to use this in our proof:



X ⊆ Y ↔ X ∩ Y = X (5)



Should this be solved by mathematical induction or somehow else? I don't know what to do about that. Sorry for bad English. That is not my first language.










share|cite|improve this question













A, B, C are subsets of a set U:



A ⊆ B → A ∩ B $nsubseteq$ C (1)



A ⊆ B ∨ A ⊆ C (2)



A ∩ C ⊆ B (3)



I have to prove that this is valid:



A ∩ B $nsubseteq$ C (4)



It is recommended to use this in our proof:



X ⊆ Y ↔ X ∩ Y = X (5)



Should this be solved by mathematical induction or somehow else? I don't know what to do about that. Sorry for bad English. That is not my first language.







elementary-set-theory proof-explanation






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asked Nov 28 at 21:17









qwerty1

11




11












  • It would be much better if you showed how much work you have done, and where exactly are you stuck. Which statements are premises, what do you want proven... so forth.
    – Bertrand Wittgenstein's Ghost
    Nov 29 at 8:40




















  • It would be much better if you showed how much work you have done, and where exactly are you stuck. Which statements are premises, what do you want proven... so forth.
    – Bertrand Wittgenstein's Ghost
    Nov 29 at 8:40


















It would be much better if you showed how much work you have done, and where exactly are you stuck. Which statements are premises, what do you want proven... so forth.
– Bertrand Wittgenstein's Ghost
Nov 29 at 8:40






It would be much better if you showed how much work you have done, and where exactly are you stuck. Which statements are premises, what do you want proven... so forth.
– Bertrand Wittgenstein's Ghost
Nov 29 at 8:40












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I'd solve it by cases on (2).
If $A subseteq B$, you're done by (1).



Suppose now that $A subseteq C$ (6).



By (6) we can say $A cap C = A$.



So by (3): $A cap C = A subseteq B$.



So $A subseteq B$ holds.



By (1): $A cap B nsubseteq C$.






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    1 Answer
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    up vote
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    down vote













    I'd solve it by cases on (2).
    If $A subseteq B$, you're done by (1).



    Suppose now that $A subseteq C$ (6).



    By (6) we can say $A cap C = A$.



    So by (3): $A cap C = A subseteq B$.



    So $A subseteq B$ holds.



    By (1): $A cap B nsubseteq C$.






    share|cite|improve this answer

























      up vote
      0
      down vote













      I'd solve it by cases on (2).
      If $A subseteq B$, you're done by (1).



      Suppose now that $A subseteq C$ (6).



      By (6) we can say $A cap C = A$.



      So by (3): $A cap C = A subseteq B$.



      So $A subseteq B$ holds.



      By (1): $A cap B nsubseteq C$.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        I'd solve it by cases on (2).
        If $A subseteq B$, you're done by (1).



        Suppose now that $A subseteq C$ (6).



        By (6) we can say $A cap C = A$.



        So by (3): $A cap C = A subseteq B$.



        So $A subseteq B$ holds.



        By (1): $A cap B nsubseteq C$.






        share|cite|improve this answer












        I'd solve it by cases on (2).
        If $A subseteq B$, you're done by (1).



        Suppose now that $A subseteq C$ (6).



        By (6) we can say $A cap C = A$.



        So by (3): $A cap C = A subseteq B$.



        So $A subseteq B$ holds.



        By (1): $A cap B nsubseteq C$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 at 21:24









        LuxGiammi

        16410




        16410






























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