Padding a 2SAT clause In http://web.mit.edu/neboat/www/6.046-fa09/rec8.pdf, I see that they pad a 2SAT clause $(x\vee y)$ to make it a 3SAT clause by writing $(x\vee y\vee p) \wedge (x\vee y\vee \neg p)$. Why doesn't $(x\vee y\vee y)$ work instead? Isn't the truth table identical?
Sometimes we insist that the three literals in a 3SAT clause belong to different variables. This ensures, for example, that a random assignment satisfies a clause with probability exactly $7/8$. The translation $x \lor y \lor y$ doesn't satisfy this condition, but the other one does.