0.00/0.07	YES
0.00/0.07	
0.00/0.07	DP problem for innermost termination.
0.00/0.07	P =
0.00/0.07	  f7#(x1, x2) -> f6#(x1, x2) 
0.00/0.07	  f6#(I0, I1) -> f5#(1, I1) 
0.00/0.07	  f5#(I2, I3) -> f3#(I2, 1) 
0.00/0.07	  f5#(I4, I5) -> f1#(I4, I5) [1 + I4 <= 1]
0.00/0.07	  f1#(I8, I9) -> f3#(I8, 0) 
0.00/0.07	  f2#(I10, I11) -> f1#(I10, I11) 
0.00/0.07	  f1#(I12, I13) -> f2#(I12, I13) [1 <= 0]
0.00/0.07	R = 
0.00/0.07	  f7(x1, x2) -> f6(x1, x2) 
0.00/0.07	  f6(I0, I1) -> f5(1, I1) 
0.00/0.07	  f5(I2, I3) -> f3(I2, 1) 
0.00/0.07	  f5(I4, I5) -> f1(I4, I5) [1 + I4 <= 1]
0.00/0.07	  f3(I6, I7) -> f4(I6, I7) [1 + I7 <= 1]
0.00/0.07	  f1(I8, I9) -> f3(I8, 0) 
0.00/0.07	  f2(I10, I11) -> f1(I10, I11) 
0.00/0.07	  f1(I12, I13) -> f2(I12, I13) [1 <= 0]
0.00/0.07	
0.00/0.07	The dependency graph for this problem is:
0.00/0.07	  0 -> 1
0.00/0.07	  1 -> 2
0.00/0.07	  2 -> 
0.00/0.07	  3 -> 4
0.00/0.07	  4 -> 
0.00/0.07	  5 -> 4
0.00/0.07	  6 -> 
0.00/0.07	Where:
0.00/0.07	  0) f7#(x1, x2) -> f6#(x1, x2) 
0.00/0.07	  1) f6#(I0, I1) -> f5#(1, I1) 
0.00/0.07	  2) f5#(I2, I3) -> f3#(I2, 1) 
0.00/0.07	  3) f5#(I4, I5) -> f1#(I4, I5) [1 + I4 <= 1]
0.00/0.07	  4) f1#(I8, I9) -> f3#(I8, 0) 
0.00/0.07	  5) f2#(I10, I11) -> f1#(I10, I11) 
0.00/0.07	  6) f1#(I12, I13) -> f2#(I12, I13) [1 <= 0]
0.00/0.07	
0.00/0.07	We have the following SCCs.
0.00/0.07	  
0.00/3.05	EOF