The operation x,y is defined by x,y = (x + 1)(y + 1) – 1 = xy + x + y. Which of these statements is false?
 x,y = y,x for all real x and y.
 x,(y + z) = (x,y) + (x,z) for all real x, y, and z.
 (x – 1),(x + 1) = (x,x) – 1 for all real x and y.
 x,0 = x for all real x.
 x,(y,z) = (x,y),z for all real x, y, and z.
 x,y = xy + x + y; y,x = yx + y + x, so A’s OK.

x,(y + z) = x(y + z) + x + (y + z) = xy + xz + x + y + z;
(x,y) + (x,z) = xy + x + y + xz + x + z = xy + xz + 2x + y + z. B is false. 
(x – 1),(x + 1) = (x – 1)(x + 1) + (x – 1) + (x + 1) = x^{2} – 1 + x – 1 + x + 1
= x^{2} + 2x – 1
(x,x) – 1 = x · x + x + x – 1 = x^{2} + 2x – 1, so C’s OK.  (x,0) = x · 0 + x + 0 = x, so D’s OK.

x,(y,z) = x(y,z) + x + (y,z) = x(yz + y + z) + x + (yz + y + z)
= xyz + xy + xz + x + yz + y + z
= xyz + xy + xz + yz + x + y + z (looks pretty symmetric!)
(x,y),z = (xy + x + y), z = xyz + xz + yz + xy + x + y + z
= xyz + xy + xz + yz + x + y + z, so E’s OK.
Statement B is the only one that’s false.