Suppose x and y are both female. We know that (x-y)+y=x. If x-y is male, then we have male+female=female, which contradicts the first rule. Therefore, x-y is female. This means that female numbers are closed under subtraction.

We know that there is at least one female number, x. Since x and x are both female, x-x=0 is female. Since 0 and x are both female, 0-x=-x is female. Since x and -x are both female, x--x=2x is female. Since 2x--x=3x, 3x is female. Since 3x--x=4x, 4x is female. Therefore, nx is female for any positive integer n. Since -x-x=-2x, -2x is female. Since -2x-x=-3x, -3x is female. Since -3x-x=-4x, -4x is female. Therefore, nx is female for any integer n. If x and y are both female, then xy is female, so R1 is female.

Suppose x and y are both female. Then, -y is also female. Therefore, x--y=x+y is female, so R2 is female.

If 0 is the only female number, then the female numbers are the multiples of 0. Otherwise, there is a female number x that is not 0. If x is positive, then there is a positive female number. If x is negative, then -x is a positive female number. Therefore, there is a positive female number. Let x be the smallest positive female number. We know that all multiples of x are female. Suppose y is not a multiple of x. Then, y=ax+b for some integers a and b, 0<b<x. Since x is the smallest positive female number, b is male. Since all multiples of x are female, ax is female. Therefore, b+ax=y is male since male+female=male. This means that the female numbers are the multiples of x.

Suppose x=10. Then, 2, 3, and 5 are all male. 2*5=10 is female, but 2*3=6 is male. Therefore, R3 could be either. 5+5=10 is female, but 3+3=6 is male. Therefore, R4 could be either.