PYEDA truthtable of functions - python-3.x

Question

Hope there is anybody who feels good with PYEDA.
I want to add fictious variables to function
Let me have f=x1, but how can I get truthtable for this function , which will have x2 too
Like truthtable for f(x1)=x1 is:
x1 f
0 0
1 1
But for f(x1,x2)=x1 is:
x1 x2 f
0 0 0
0 1 0
1 0 1
1 1 1
But I will get first table, pyeda will simplify x1&(x2|~x2) to x1 automatically. How can I add this x2?
def calcFunction(function, i):
#here is is point with dimension-size 4
function=function.restrict({x4:i[3]})
function = function.restrict({x3:i[2]})
function = function.restrict({x2:i[1]})
function = function.restrict({x1:i[0]})
if function.satisfy_one() is not None:
return 1
return 0
Here is my algo to fix it, I am calculating func in each point manually, where function can containt 1-4 variables and I am calculating for all point and combinations of x1...x4.

Answer 1

I'm not sure I understand the question as asked, but you might want to try the expression simplify method.
For example:
In [1]: f = (X[1] & X[2]) | (X[3] | X[4] | ~X[3])
In [2]: expr2truthtable(f)
Out[2]:
x[4] x[3] x[2] x[1]
0 0 0 0 : 1
0 0 0 1 : 1
0 0 1 0 : 1
0 0 1 1 : 1
0 1 0 0 : 1
0 1 0 1 : 1
0 1 1 0 : 1
0 1 1 1 : 1
1 0 0 0 : 1
1 0 0 1 : 1
1 0 1 0 : 1
1 0 1 1 : 1
1 1 0 0 : 1
1 1 0 1 : 1
1 1 1 0 : 1
1 1 1 1 : 1
In [3]: f = f.simplify()
In [4]: f
Out[4]: 1
In [5]: expr2truthtable(f)
Out[5]: 1