0.59/0.62	MAYBE
0.59/0.62	
0.59/0.62	DP problem for innermost termination.
0.59/0.62	P =
0.59/0.62	  f4#(x1) -> f3#(x1) 
0.59/0.62	  f3#(I0) -> f1#(I0) 
0.59/0.62	  f2#(I1) -> f1#(I1) 
0.59/0.62	  f1#(I2) -> f2#(1 - I2) [I2 <= 1 /\ 0 <= I2]
0.59/0.62	R = 
0.59/0.62	  f4(x1) -> f3(x1) 
0.59/0.62	  f3(I0) -> f1(I0) 
0.59/0.62	  f2(I1) -> f1(I1) 
0.59/0.62	  f1(I2) -> f2(1 - I2) [I2 <= 1 /\ 0 <= I2]
0.59/0.62	
0.59/0.62	The dependency graph for this problem is:
0.59/0.62	  0 -> 1
0.59/0.62	  1 -> 3
0.59/0.62	  2 -> 3
0.59/0.62	  3 -> 2
0.59/0.62	Where:
0.59/0.62	  0) f4#(x1) -> f3#(x1) 
0.59/0.62	  1) f3#(I0) -> f1#(I0) 
0.59/0.62	  2) f2#(I1) -> f1#(I1) 
0.59/0.62	  3) f1#(I2) -> f2#(1 - I2) [I2 <= 1 /\ 0 <= I2]
0.59/0.62	
0.59/0.62	We have the following SCCs.
0.59/0.62	  { 2, 3 }
0.59/0.62	
0.59/0.62	DP problem for innermost termination.
0.59/0.62	P =
0.59/0.62	  f2#(I1) -> f1#(I1) 
0.59/0.62	  f1#(I2) -> f2#(1 - I2) [I2 <= 1 /\ 0 <= I2]
0.59/0.62	R = 
0.59/0.62	  f4(x1) -> f3(x1) 
0.59/0.62	  f3(I0) -> f1(I0) 
0.59/0.62	  f2(I1) -> f1(I1) 
0.59/0.62	  f1(I2) -> f2(1 - I2) [I2 <= 1 /\ 0 <= I2]
0.59/0.62	
0.59/3.60	EOF