0.89/0.89	MAYBE
0.89/0.89	
0.89/0.89	DP problem for innermost termination.
0.89/0.89	P =
0.89/0.89	  f5#(x1, x2) -> f4#(x1, x2) 
0.89/0.89	  f4#(I0, I1) -> f1#(I0, I1) [1 <= I0 /\ 1 + I0 + I1 <= 0]
0.89/0.89	  f3#(I2, I3) -> f1#(I2, I3) 
0.89/0.89	  f1#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I4]
0.89/0.89	R = 
0.89/0.89	  f5(x1, x2) -> f4(x1, x2) 
0.89/0.89	  f4(I0, I1) -> f1(I0, I1) [1 <= I0 /\ 1 + I0 + I1 <= 0]
0.89/0.89	  f3(I2, I3) -> f1(I2, I3) 
0.89/0.89	  f1(I4, I5) -> f3(I4 - I5, I5) [0 <= I4]
0.89/0.89	  f1(I6, I7) -> f2(I6, I7) [1 + I6 <= 0]
0.89/0.89	
0.89/0.89	The dependency graph for this problem is:
0.89/0.89	  0 -> 1
0.89/0.89	  1 -> 3
0.89/0.89	  2 -> 3
0.89/0.89	  3 -> 2
0.89/0.89	Where:
0.89/0.89	  0) f5#(x1, x2) -> f4#(x1, x2) 
0.89/0.89	  1) f4#(I0, I1) -> f1#(I0, I1) [1 <= I0 /\ 1 + I0 + I1 <= 0]
0.89/0.89	  2) f3#(I2, I3) -> f1#(I2, I3) 
0.89/0.89	  3) f1#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I4]
0.89/0.89	
0.89/0.89	We have the following SCCs.
0.89/0.89	  { 2, 3 }
0.89/0.89	
0.89/0.89	DP problem for innermost termination.
0.89/0.89	P =
0.89/0.89	  f3#(I2, I3) -> f1#(I2, I3) 
0.89/0.89	  f1#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I4]
0.89/0.89	R = 
0.89/0.89	  f5(x1, x2) -> f4(x1, x2) 
0.89/0.89	  f4(I0, I1) -> f1(I0, I1) [1 <= I0 /\ 1 + I0 + I1 <= 0]
0.89/0.89	  f3(I2, I3) -> f1(I2, I3) 
0.89/0.89	  f1(I4, I5) -> f3(I4 - I5, I5) [0 <= I4]
0.89/0.89	  f1(I6, I7) -> f2(I6, I7) [1 + I6 <= 0]
0.89/0.89	
0.89/3.87	EOF