On your third line, you have added a negation in front of y2 that is not present on the second line. So if I follow the formula from the second line:

(!x|(!y1|y2))& ((!y1|y2)->x)

(!x | !y1 | y2)&((y1&!y2) | x)

(!x | !y1 | y2) & (x | y1) & (x | !y2)

Other than the extra negation in front of y2, there seems to be no other error.