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something is. Thanks for explanation, too..

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9 thoughts on “ A.B.C.
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  5. Gardat says:
    My initial answer suggested that this was an axiom - but on reflection: Assuming that a,b,c are all real numbers (i.e. not complex). The definition of A > B is that A-B is a positive real number. Since a>b then a-b is a positive real number (call.
  6. Voodoolabar says:
    Feb 24,  · Let A,B and C be sets. Prove that (A-B)-C=(A-C)-(B-C). Attempted solution: Suppose x \\in (A-B)-C. Since x \\in (A-B)-C this means that x \\in A but x \\notin B and x \\notin C. I'm not sure how to show how these two statements are equal.
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  9. Gardarg says:
    Sep 30,  · Let us first define some basic formulas of boolean algebra that we're gonna need for the solution 1. (X+X)=X 2. (X.X)=X 3. (X+1)=1 4. 1.X=X Now let us get to the problem (A+B)(B+C)(A+C) =(AB+AC+BC+BB)(A+C) {Multiplying the first two terms} =(AB+AC.

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