@acer 

Thanks again for your helps and useful comments.

@acer 

What should I do If I want to generalize your procedure to any input F and any monomial ordering T?

@acer 

Thank you so much for your solution this is OK.

Sincerely yours

@Carl Love 

Thanks, No I need a function or procedure for computing in polynomial quotient rings. Let I  be a homogeneous polynomial ideal of degree d (e.g. I=<x-y>) and R=K[x,y,z]. So R/I = K[x,y,z] / I ------> K[y,z] or K[x,z]. Also the polynomial ideal [x^2+y^2+z^2] changes into [2y^2+z^2] in R/I. I dont know how do automatically this in Maple.

Thanks again.

@Christian Wolinski 

Thank you so much for your answer.

@Carl Love 

Thank you so much for your efficient method.

Sincerely yours

@acer 

Thank you so much. So, using your comparison performance you suggest rtable_scanblock. Is this true?

You have to compute a Grobner basis w.r.t. a lex ordering not tdeg. Also, you could use the solve command for this.

@acer Thank you so much. I am trying to do it.

@Kitonum Oh my God!! this is excellent. Thank you so much.

@Kitonum Thank you so much. This is perfect!

@Kitonum 
Thanks again for this exellent method. Your method is appropriate for this system which arise when "n=3" and in your method I see that you use from S1 and S2 contained three polynomial. How can I use from this method automatically for any arbitrary n? 

@Kitonum 

Thank you so much for your replying. It seems that the pervious method is not appropriate for the following list. (I know that this is different list). In this list I want to subs:

E[k,k]= -1 when there is "(a+2)*x[k,k]+E[k,k]" as a part of polynomial in the following list

and 

E[k,k]= 1 when there is "a*x[k,k]+E[k,k]" as a part of polynomial in the following list.

For example if

B=[(a*x[1, 1] + E[1, 1])*(E[1, 1] + (a + 2)*x[1, 1]) + (a*x[2, 1] + E[2, 1])*(E[1, 2] + (a + 2)*x[1, 2]) + (a*x[3, 1] + E[3, 1])*(E[1, 3] + (a + 2)*x[1, 3])] 

then after subsituation we have 

B=[(a*x[1, 1] + 1)*(-1 + (a + 2)*x[1, 1]) + (a*x[2, 1] + E[2, 1])*(E[1, 2] + (a + 2)*x[1, 2]) + (a*x[3, 1] + E[3, 1])*(E[1, 3] + (a + 2)*x[1, 3])]

Could you please help me?

A=[(a*x[1, 1] + E[1, 1])*(E[1, 1] + (a + 2)*x[1, 1]) + (a*x[2, 1] + E[2, 1])*(E[1, 2] + (a + 2)*x[1, 2]) + (a*x[3, 1] + E[3, 1])*(E[1, 3] + (a + 2)*x[1, 3]), (a*x[1, 2] + E[1, 2])*(E[1, 1] + (a + 2)*x[1, 1]) + (a*x[2, 2] + E[2, 2])*(E[1, 2] + (a + 2)*x[1, 2]) + (a*x[3, 2] + E[3, 2])*(E[1, 3] + (a + 2)*x[1, 3]), (a*x[1, 3] + E[1, 3])*(E[1, 1] + (a + 2)*x[1, 1]) + (a*x[2, 3] + E[2, 3])*(E[1, 2] + (a + 2)*x[1, 2]) + (a*x[3, 3] + E[3, 3])*(E[1, 3] + (a + 2)*x[1, 3]), (a*x[1, 1] + E[1, 1])*(E[2, 1] + (a + 2)*x[2, 1]) + (a*x[2, 1] + E[2, 1])*(E[2, 2] + (a + 2)*x[2, 2]) + (a*x[3, 1] + E[3, 1])*(E[2, 3] + (a + 2)*x[2, 3]), (a*x[1, 2] + E[1, 2])*(E[2, 1] + (a + 2)*x[2, 1]) + (a*x[2, 2] + E[2, 2])*(E[2, 2] + (a + 2)*x[2, 2]) + (a*x[3, 2] + E[3, 2])*(E[2, 3] + (a + 2)*x[2, 3]), (a*x[1, 3] + E[1, 3])*(E[2, 1] + (a + 2)*x[2, 1]) + (a*x[2, 3] + E[2, 3])*(E[2, 2] + (a + 2)*x[2, 2]) + (a*x[3, 3] + E[3, 3])*(E[2, 3] + (a + 2)*x[2, 3]), (a*x[1, 1] + E[1, 1])*(E[3, 1] + (a + 2)*x[3, 1]) + (a*x[2, 1] + E[2, 1])*(E[3, 2] + (a + 2)*x[3, 2]) + (a*x[3, 1] + E[3, 1])*(E[3, 3] + (a + 2)*x[3, 3]), (a*x[1, 2] + E[1, 2])*(E[3, 1] + (a + 2)*x[3, 1]) + (a*x[2, 2] + E[2, 2])*(E[3, 2] + (a + 2)*x[3, 2]) + (a*x[3, 2] + E[3, 2])*(E[3, 3] + (a + 2)*x[3, 3]), (a*x[1, 3] + E[1, 3])*(E[3, 1] + (a + 2)*x[3, 1]) + (a*x[2, 3] + E[2, 3])*(E[3, 2] + (a + 2)*x[3, 2]) + (a*x[3, 3] + E[3, 3])*(E[3, 3] + (a + 2)*x[3, 3])]

@vv  Thank you so much. Now, it is good.!! How select this as the best answer?

@vv Thanks again. I think my question is obvious and my example is completely appropriate. I need a function which receives two parametric monomials (and also a list of variables) and returns true or false;

its output is true if these monomials have not any common variables and return false if there is at least one common variable.