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I want to obtain a solution for a system of five inequalities. For instance,

Reduce[A*B + C + D > 0 && A > 0 && A + B > 1 && A*B > 2 && C > 0 && 
  D < 1, {A, B}]

Typing this into Mathematica, I obtain the following output:

C > 0 && ((D <= -2 - C && A > 0 && 
     B > (-C - D)/A) || (-2 - C < D < 1 && A > 0 && B > 2/A))

Then my friend asked me this question: is this solution a necessary and sufficient condition for fulfilling the inequalities given above?

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    $\begingroup$ FYI: C and D are reserved Symbols and should not be used like this; in fact it is better to avoid starting any user Symbols with capital letters. $\endgroup$ – Mr.Wizard May 19 '15 at 3:08
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Yes, it is documented that the result of Reduce[expr, vars] always describes exactly the same mathematical set as expr, i.e. the result of the reduction is equivalent to the original system.

Another way to state the above is

In[1]:= s1 = c > 0 && ((d <= -2 - c && a > 0 && b > (-c - d)/a) || 
               (-2 - c < d < 1 && a > 0 && b > 2/a));
        s2 = a*b + c + d > 0 && a > 0 && a + b > 1 && a*b > 2 && c > 0 && d < 1;
        Resolve[ForAll[{a, b, c, d}, Equivalent[s1, s2]]]

Out[3]= True
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