I was looking to see if someone could let me know if I am doing this problem right.
I got a 13/14 and have one more attempt. Originally I answered true but I attempted to solve the question a different way and got a new answer.
This is the question
This is my new solution
You should be careful about the order of every step
I want to know if the symbol " ’ " is not. If it is yes.
Then y’ is True. And the following is the answer F.
 pry(main)> w = true
 pry(main)> x = true
 pry(main)> y = false
 pry(main)> (((w || x) && (x && !y)) ||((w || x) && (!w || !y))) && !((x && !y) && (!w || !y ))