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  • tux

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

Mathematics
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Now, A∩(B-C)

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Other answers:

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