How to remove left recursion (source code Python)

l = [ 'A'] #lhs of grammar
#R = [ ['A','a'],['B']] #rhs of grammar
#R = [ [ 'A','a'], ['A','b'],['A','c'], ['B'],['C'],['D'] ]
R = [ ['A','+','A'],[ 'A','*','A' ],['a'],['b'],['c'] ]
nl = [] #new left
nR = [] #new Right
nRightWithRecursion = [] #new Right side Grammar with Recursion, has Recursion
fR = [nR,nRightWithRecursion] #to print, list for new Right

lenR = len(R) #length of right side
lenL = len(l)
lenInnerList = 0

for i in range(lenR):#iright grammar
lenInnerList = len(R[i])

for j in range(lenL): # at inner grammar, to compare left with right grammar
#print [ 'R', R,'i',i,'j',j ] #debug print
if(l[j] == R[i][j]): #same symbol on left, right inner grammar
R[i].pop(0)
nRightWithRecursion.append( R[i] )
else:
nR.append( R[i] ) #take where there is no left recursion
break #not equal don't do anything, move to next or grammar

#add new variables where there is no left recursion
#-------------------------------------------------
nVar = 1 # new variable for left and right

lenNewRight = len(nR)
nl.append( [ l[0] ]) # adding a list, previous left

for i in range(lenNewRight):
nR[i].append(nVar)

lenNewRightWithRec = len(nRightWithRecursion)
nl.append( [ nVar ]) # add next symbol, new left symbol, new grammar, contains right recursion's lhs

for i in range(lenNewRightWithRec):
nRightWithRecursion[i].append( nVar)

nRightWithRecursion.append(['e']) #empty string at converted right recursion

#-------------------------------------------------




# print R
# print nR
# print nRightWithRecursion

print [nl[0],"===>", fR[0] ]
print [nl[1],"===>", fR[1] ]