tux
A∩(B-C)=(A∩B)-(A∩C). Using algebraic proof I got (A∩B)∩(A-C).
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sauravshakya
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|dw:1346329193067:dw|
sauravshakya
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|dw:1346329377564:dw|
sauravshakya
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Now, A∩(B-C)
sauravshakya
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|dw:1346329433857:dw|
sauravshakya
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|dw:1346329493645:dw|Similarly, proceed step by step u will get
sauravshakya
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Thus, proved
tux
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My problem is I need to prove it using set laws (commutative, associative ...)
Mikael
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Translate to set algebra \[\cup \rightarrow +\]
|dw:1346351911567:dw|
Mikael
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|dw:1346351955786:dw|
Mikael
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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
Mikael
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Also (forgoT)|dw:1346352053975:dw|
Mikael
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|dw:1346352107065:dw|
Mikael
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|dw:1346352172712:dw|
Mikael
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Whatever form is more convenient for the right hand side expression
Mikael
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USe also |dw:1346352282882:dw|
tux
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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)
tux
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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).
Mikael
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I suggest u try boolean algebra
Mikael
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In the representation I have showed above all is solved easy
Mikael
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shown (typo)
tux
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I am not allowed to use complement definition 1-A
farmdawgnation
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@IAmCool Please do not go into other's threads asking for help on your question, it's considered spam.
zzr0ck3r
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@sauravshakya @tux @Mikael ....I think you guys were making this way to hard.
sauravshakya
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?
zzr0ck3r
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