0.89/0.94	MAYBE
0.89/0.94	
0.89/0.94	DP problem for innermost termination.
0.89/0.94	P =
0.89/0.94	  f4#(x1, x2) -> f3#(x1, x2) 
0.89/0.94	  f3#(I0, I1) -> f1#(I0, I1) 
0.89/0.94	  f2#(I2, I3) -> f1#(I2, I3) 
0.89/0.94	  f1#(I4, I5) -> f2#(I4 - I5, 1 + I5) [1 <= I4]
0.89/0.94	R = 
0.89/0.94	  f4(x1, x2) -> f3(x1, x2) 
0.89/0.94	  f3(I0, I1) -> f1(I0, I1) 
0.89/0.94	  f2(I2, I3) -> f1(I2, I3) 
0.89/0.94	  f1(I4, I5) -> f2(I4 - I5, 1 + I5) [1 <= I4]
0.89/0.94	
0.89/0.94	The dependency graph for this problem is:
0.89/0.94	  0 -> 1
0.89/0.94	  1 -> 3
0.89/0.94	  2 -> 3
0.89/0.94	  3 -> 2
0.89/0.94	Where:
0.89/0.94	  0) f4#(x1, x2) -> f3#(x1, x2) 
0.89/0.94	  1) f3#(I0, I1) -> f1#(I0, I1) 
0.89/0.94	  2) f2#(I2, I3) -> f1#(I2, I3) 
0.89/0.94	  3) f1#(I4, I5) -> f2#(I4 - I5, 1 + I5) [1 <= I4]
0.89/0.94	
0.89/0.94	We have the following SCCs.
0.89/0.94	  { 2, 3 }
0.89/0.94	
0.89/0.94	DP problem for innermost termination.
0.89/0.94	P =
0.89/0.94	  f2#(I2, I3) -> f1#(I2, I3) 
0.89/0.94	  f1#(I4, I5) -> f2#(I4 - I5, 1 + I5) [1 <= I4]
0.89/0.94	R = 
0.89/0.94	  f4(x1, x2) -> f3(x1, x2) 
0.89/0.94	  f3(I0, I1) -> f1(I0, I1) 
0.89/0.94	  f2(I2, I3) -> f1(I2, I3) 
0.89/0.94	  f1(I4, I5) -> f2(I4 - I5, 1 + I5) [1 <= I4]
0.89/0.94	
0.89/3.92	EOF