
In this DPLL algorithm, is that possible to set another value to x0 and x1. 
When i tried, 
{x0, ¬b}, {x0, ¬c}, {b, c, ¬x0} 
{x0, ¬x1}, {x1, ¬a, ¬x0},
 {a, ¬x1} {x1, ¬b}
set x1 = true, 
{x0, ¬b}, {x0, ¬c}, {b, c, ¬x0} 
{x0}
 {a} 
a = true, x0 = true, b = c = true or b = false or c = true
Is that a valid solution ? As per my understanding we should not get anything like {} which will be a conflict. When i gave these values i did not get any conflicts