document.write( "Question 626372: let A,B,andC be the sets such that AunionB=AunionCandAintersectionB=AintersectionC.show that B=C. \n" ); document.write( "
Algebra.Com's Answer #394241 by Edwin McCravy(20060)\"\" \"About 
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document.write( "Assume for contradiction that B≠C \r\n" );
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document.write( "Then either B⊈C or C⊈B \r\n" );
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document.write( "We only need to disprove one of these since after we have \r\n" );
document.write( "disproved one of them, we can disprove the other just by \r\n" );
document.write( "swapping the roles of B and C.\r\n" );
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document.write( "We will assume B⊈C\r\n" );
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document.write( "Then ∃x such that x∈B and x∉C\r\n" );
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document.write( "The either x∈A or x∉A\r\n" );
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document.write( "Case 1: x∈A.  Then since x∈B, x∈A⋂B. But\r\n" );
document.write( "since x∉C, x∉A⋂C. Therefore A⋂B≠A⋂C, a\r\n" );
document.write( "contadiction since A⋂B=A⋂C is given. So case 1 is disproved.\r\n" );
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document.write( "Case 2: x∉A. Then since x∈B, x∈A⋃B. But\r\n" );
document.write( "since x∉C, x∉A⋃C. Therefore A⋃B≠A⋃C, a\r\n" );
document.write( "contadiction, since A⋃B=A⋃C is given. So case 2 is disproved.\r\n" );
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document.write( "Therefore B⊈C is false and B⊆C is true.\r\n" );
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document.write( "By swapping the roles of B and C in the above, C⊈B is false \r\n" );
document.write( "and C⊆B is true.\r\n" );
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document.write( "Therefore B=C.\r\n" );
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document.write( "Edwin
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