# Some question on “Reduce”

Is this my mistake?

I am trying to solve

   Assuming[0 < a < 1 && 0 < b < 1 && 0 < c < 1 && 0 < n < 2 && Q > 0 &&
0 < p <= 1 && Element[a | b | c | n | p | Q, Reals],
Reduce[2*a*b*n*c*Q - 3*a*b*n*p*c*Q + a*b*n*p^2*c*Q -
4*a^2*b*n*p*Q^2 - 4*a*b^2*n*p*Q^2 + 6*a^2*b*n*p^2*Q^2 +
6*a*b^2*n*p^2*Q^2 - 2*a^2*b*n*p^3*Q^2 - 2*a*b^2*n*p^3*Q^2 +
a^2*b*c*Q^2 + a*b^2*c*Q^2 - 2*a^2*b*p*c*Q^2 - 2*a*b^2*p*c*Q^2 ==
0]]


However, the answer provided doesn't seem to be fit with my assumption. For instance, I clearly state that $0<p<=1$. But in the answer, it give me $p=2$, and also $p=0$. How do I filter out these unwanted values?

The program returns,

     (Q == 0 && (p == 1 ||
p == 2)) || ((-c + 2 c p + 4 n p - 6 n p^2 + 2 n p^3) Q != 0 &&
a == (2 c n - 3 c n p + c n p^2 + b c Q - 2 b c p Q - 4 b n p Q +
6 b n p^2 Q -
2 b n p^3 Q)/((-c + 2 c p + 4 n p - 6 n p^2 +
2 n p^3) Q)) || (p == 1/2 && n == 0 && Q != 0) || (Q == 0 &&
2 - 3 p + p^2 != 0 && n == 0) || (Q == 0 &&
n (2 - 3 p + p^2) != 0 && c == 0) || ((p == 1 || p == 2) &&
c == 0 && Q != 0) || (n == 0 && -1 + 2 p != 0 && c == 0 &&
2 p Q - 3 p^2 Q + p^3 Q != 0) || (p == 0 && n == 0 && c == 0 &&
Q != 0) || (Q != 0 && p == 0 && 2 n != 0 && c == 0) || a == 0 ||
b == 0 || Q == 0

-

Reduce has its own syntax for specifying constrains and domains.
Reduce[2*a*b*n*c*Q - 3*a*b*n*p*c*Q + a*b*n*p^2*c*Q -