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tux

  • 2 years ago

A∩(B-C)=(A∩B)-(A∩C). Using algebraic proof I got (A∩B)∩(A-C).

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  1. sauravshakya
    • 2 years ago
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    |dw:1346329193067:dw|

  2. sauravshakya
    • 2 years ago
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    |dw:1346329377564:dw|

  3. sauravshakya
    • 2 years ago
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    Now, A∩(B-C)

  4. sauravshakya
    • 2 years ago
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    |dw:1346329433857:dw|

  5. sauravshakya
    • 2 years ago
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    |dw:1346329493645:dw|Similarly, proceed step by step u will get

  6. sauravshakya
    • 2 years ago
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    Thus, proved

  7. tux
    • 2 years ago
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    My problem is I need to prove it using set laws (commutative, associative ...)

  8. Mikael
    • 2 years ago
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    Translate to set algebra \[\cup \rightarrow +\] |dw:1346351911567:dw|

  9. Mikael
    • 2 years ago
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    |dw:1346351955786:dw|

  10. Mikael
    • 2 years ago
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    translate to this notation your problem, THEN open all parantheses, THEN gather like terms, THEN translate back to Set theory notation

  11. Mikael
    • 2 years ago
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    Also (forgoT)|dw:1346352053975:dw|

  12. Mikael
    • 2 years ago
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    |dw:1346352107065:dw|

  13. Mikael
    • 2 years ago
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    |dw:1346352172712:dw|

  14. Mikael
    • 2 years ago
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    Whatever form is more convenient for the right hand side expression

  15. Mikael
    • 2 years ago
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    USe also |dw:1346352282882:dw|

  16. tux
    • 2 years ago
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    B-C rewritten as B∩C^c A∩(B∩C^c) A rewritten as A∩A A∩A∩B∩C^c Associative law (A∩B)∩(A∩C^c) My result: (A∩B)∩(A-C)

  17. tux
    • 2 years ago
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    @sauravshakya We start with left side A∩(B-C) and must prove (A∩B)-(A∩C) Solution is (A∩B)-(A∩C) which must be proved As a wrong result I got (A∩B)∩(A-C).

  18. Mikael
    • 2 years ago
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    I suggest u try boolean algebra

  19. Mikael
    • 2 years ago
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    In the representation I have showed above all is solved easy

  20. Mikael
    • 2 years ago
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    shown (typo)

  21. tux
    • 2 years ago
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    I am not allowed to use complement definition 1-A

  22. farmdawgnation
    • 2 years ago
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    @IAmCool Please do not go into other's threads asking for help on your question, it's considered spam.

  23. zzr0ck3r
    • 2 years ago
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    @sauravshakya @tux @Mikael ....I think you guys were making this way to hard.

  24. sauravshakya
    • 2 years ago
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    ?

  25. zzr0ck3r
    • 2 years ago
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