How to "read" the solution of an inequality?

Question

The solution of fun[x,y]>0 leads to the following output

 fun[x_, y_] = (1/16)*(1 - 6*x*y + (-1 + x)*x*y)*(1 + x*(-1 + 6*y) + 
     x*(-1 + x + y - x*y))*(-1 + 2*x - 
     x^2 + (x + (-1 + x)*x*(-1 + y) - 6*x*y)*(6*x*y + (x - x^2)*y)); 
Reduce[fun[x, y] > 0 && 0 < x < 1 && 0 < y < 1, y]

$$\color{blue}{\left(\frac{1}{2} \left(7-3 \sqrt{5}\right)<x\leq 2-\sqrt{2}\land -\frac{1}{x^2-7 x}<y<1\right) }\lor \color{red}{ \left(2-\sqrt{2}<x<1 \land \left(\frac{x-2}{2 (x-7)}-\frac{1}{2} \sqrt{\frac{x^4-4 x^3+8 x-4}{(x-7)^2 x^2}}<y<\frac{1}{2} \sqrt{\frac{x^4-4 x^3+8 x-4}{(x-7)^2 x^2}}+\frac{x-2}{2 (x-7)}\lor -\frac{1}{x^2-7 x}<y<1\right)\right)}$$

The way I understand this is:

1. blue and red are two solutions
2. when $x$ lies in range $\frac{1}{2} \left(7-3 \sqrt{5}\right)<x\leq 2-\sqrt{2}$, $y$ can take values in the range $\color{green}{-\frac{1}{x^2-7 x}<y<1}$
3. when $x$ lies in range $2-\sqrt{2}<x<1$, $y$ can take values in the range $\frac{x-2}{2 (x-7)}-\frac{1}{2} \sqrt{\frac{x^4-4 x^3+8 x-4}{(x-7)^2 x^2}}<y<\frac{1}{2} \sqrt{\frac{x^4-4 x^3+8 x-4}{(x-7)^2 x^2}}+\frac{x-2}{2 (x-7)}$ $\textbf{or}$ $\color{green}{-\frac{1}{x^2-7 x}<y<1}$

Is Mathematica telling me that $y$ exists in range $-\frac{1}{x^2-7 x}<y<1$ for two different ranges of $x$?

Doesn't 2 and 3 imply $-\frac{1}{x^2-7 x}<y<1$ for $\frac{1}{2} \left(7-3 \sqrt{5}\right)<x\leq1$?

Answer 1

In Make Reduce produce nicer output @chuy presented a comand line to make a nice looking output ot results of Reduce . I here give the form that also worked in version 8.0.

fun[x_, y_] = (1/16)*(1 - 6*x*y + (-1 + x)*x*y)*(1 + x*(-1 + 6*y) + 
 x*(-1 + x + y - x*y))*(-1 + 2*x - 
 x^2 + (x + (-1 + x)*x*(-1 + y) - 6*x*y)*(6*x*y + (x - x^2)*y));

red = Reduce[fun[x, y] > 0 && 0 < x < 1 && 0 < y < 1, y];

TraditionalForm[
red //. Or -> 
Composition[(Column[#, Right, Background -> {{White, LightGray}}, 
   Frame -> All] &), List]]

TraditionalForm[
 N[red] //. 
 Or -> Composition[(Column[#, Right, 
   Background -> {{White, LightGray}}, Frame -> All] &), List]]

Answer 2

How about

FullSimplify[(1/2 (7 - 3 Sqrt[5]) < x <= 
 2 - Sqrt[2] && -(1/(-7 x + x^2)) < y < 1) || (2 - Sqrt[2] < x < 
 1 && ((-2 + x)/(2 (-7 + x)) - 
 1/2 Sqrt[(-4 + 8 x - 4 x^3 + x^4)/((-7 + x)^2 x^2)] < 
 y < (-2 + x)/(2 (-7 + x)) + 
 1/2 Sqrt[(-4 + 8 x - 4 x^3 + 
         x^4)/((-7 + x)^2 x^2)] || -(1/(-7 x + x^2)) < y < 1)), 
 Assumptions -> 0 < x < 1 && 0 < y < 1]

1 + (-7 + x) x y < 0 || (-2 + x + Sqrt[-4 + 8 x - 4 x^3 + x^4]/x)/( 2 (-7 + x)) < y < (-2 + x - Sqrt[-4 + 8 x - 4 x^3 + x^4]/x)/( 2 (-7 + x))

? Addition.

RegionPlot[1 + (-7 + x) x y < 0 || (-2 + x + Sqrt[-4 + 8 x - 4 x^3 + x^4]/x)/
(2 (-7 + x)) < y < (-2 + x - Sqrt[-4 + 8 x - 4 x^3 + x^4]/x)/(2 (-7 + x)), 
{x, 0, 1}, {y, 0, 1}, PlotPoints -> 50]

RegionPlot[1 + (-7 + x) x y < 0, {x, 0, 1}, {y, 0, 1},PlotPoints -> 50]

RegionPlot[(-2 + x + Sqrt[-4 + 8 x - 4 x^3 + x^4]/x)/(2 (-7 + x)) < y < (-2 + x - Sqrt[-4 + 8 x - 4 x^3 + x^4]/x)/(2 (-7 + x)), {x, 0, 1}, {y, 0, 1}, PlotPoints -> 50]