Example (Cells)

ContourPlot[{(x^2)+(y^2)-4==0,(x-1)(y-1)-1==0},{x,-5,5},{y,-5,5}]
RegionPlot[(x^2)+(y^2)-4>0&&(x-1)(y-1)-1>0,{x,-5,5},{y,-5,5}]
RegionPlot[(x^2)+(y^2)-4<0&&(x-1)(y-1)-1>0,{x,-5,5},{y,-5,5}]
RegionPlot[(x^2)+(y^2)-4>0&&(x-1)(y-1)-1<0,{x,-5,5},{y,-5,5}]
RegionPlot[(x^2)+(y^2)-4<0&&(x-1)(y-1)-1<0,{x,-5,5},{y,-5,5}]

The Tacnode Example

p[x_,y_]:=2(x^4)-3(x^2)y+(y^4)-2(y^3)+(y^2);
ContourPlot[p[x,y]==0,{x,-2,2},{y,0,2.5}]

CAD with respect to x,y

Reduce[Discriminant[p[x,y],y]==0,x,Reals]//N

ContourPlot[{p[x,y]==0,x==0,x==-0.2365557162041041`,x==0.2365557162041041`,x==-1.4969203224061072`,x==1.4969203224061072`},{x,-2,2},{y,0,2.5}]

CAD with respect to y, x

Reduce[Discriminant[p[x,y],x]==0,y,Reals]//N

ContourPlot[{p[x,y]==0,y==0,y==1,y==-0.060660171779821415`,y==2.0606601717798214`},{x,-2,2},{y,-0.5,2.5}]

Solution (Region of truth)

ContourPlot[{p[x,y]==0,y==0,y==1,y==-0.060660171779821415`,y==2.0606601717798214`,x==0,x==-0.2365557162041041`,x==0.2365557162041041`,x==-1.4969203224061072`,x==1.4969203224061072`},{x,-2,2},{y,-0.5,2.5}]

with CAD

CylindricalDecomposition[2(x^4)-3(x^2)y+(y^4)-2(y^3)+(y^2)==0,{x,y}]//N

(x==-1.4969203224061072`&&y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,1])||(-1.4969203224061072`<x<-0.2365557162041041`&&(y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,1]||y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,2]))||(x==-0.2365557162041041`&&(y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,1]||y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,2]||y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,4]))||(-0.2365557162041041`<x<0.`&&(y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,1]||y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,2]||y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,3]||y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,4]))||(x==0.`&&(y==0.`||y==1.`))||(0.`<x<0.2365557162041041`&&(y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,1]||y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,2]||y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,3]||y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,4]))||(x==0.2365557162041041`&&(y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,1]||y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,2]||y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,4]))||(0.2365557162041041`<x<1.4969203224061072`&&(y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,1]||y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,2]))||(x==1.4969203224061072`&&y==Root[2 (x^4)-3 (x^2) #1+((#1)^2)-2 ((#1)^3)+((#1)^4)&,1])

Circle

p[x_,y_]:=x^2+y^2-1;
ContourPlot[p[x,y]==0 ,{x,-2,2},{y,-2,2}]

CylindricalDecomposition[{x^2-y^2-1==0,y≥0},{x,y}]//N

Discriminant[p[x,y],y]

Reduce[Discriminant[p[x,y],y]==0,x,Reals]//N

ContourPlot[p[x,y]==0,{x,-1,1},{y,0,1}]

Solve[x^2-y^2-1==0,y]

Sample points in x

Solve[p[x,y]==0,x]

There are 2 real roots, x=±1, therefore the sample points are {-3/2,-1,0,1,3/2}.

Sample points in (R^2)

Solve[p[x,y]== 0,y]

Sample points are (-1,-1),(-1,1),(-1,0),(0,-(Sqrt[-1+(x^2)])-1),(0,-(Sqrt[-1+(x^2)])),(0,0),(0,(Sqrt[-1+(x^2)])),(0,(Sqrt[-1+(x^2)])+1),(1,-1),(1,0),(1,1).

Truth values for x^2 - y^2 = 1 ∧ y >= 0

True for (-1,0),(0,(Sqrt[-1+(x^2)])),(1,0)

Therefore for - 1 < x < 1 => y=(Sqrt[-1+(x^2)]), the solution od CAD.

Other example

Plot[{x^2-2x,x^2-4x+3},{x,-0.2,3.5},PlotRange-> Automatic]

Solve[x^2-2x==0,x]

Solve[x^2-4x+3==0,x]

Both functions has to be >= 0.
Therefore , we have following real roots {0, 1, 2, 3}.
Sample points : {-1, 0, 1/2, 1, 3/2, 2, 5/2, 3, 4}

For quantified formulas the sample points determine whether the formulas are true or false.
If we have: ∃x∈R: (x^2-2x>=0)∧(x^2-4x+3>=0).
For x=0, it is true.
For x=1, it is false.

If we have: ∀x∈R: (x^2-2x>=0)∧(x^2-4x+3>=0), therefore it is false, because we have to implement all sample points.

CylindricalDecomposition[{x^2-2x≥ 0,x^2-4x+3≥ 0},{x,y}]//N