0.00/0.25	MAYBE
0.00/0.25	
0.00/0.25	DP problem for innermost termination.
0.00/0.25	P =
0.00/0.25	  f6#(x1) -> f5#(x1) 
0.00/0.25	  f5#(I0) -> f1#(I0) 
0.00/0.25	  f4#(I1) -> f3#(I1) 
0.00/0.25	  f3#(I2) -> f4#(I2) 
0.00/0.25	  f2#(I3) -> f1#(I3) 
0.00/0.25	  f1#(I4) -> f2#(I4) [I4 <= I4 /\ I4 <= I4]
0.00/0.25	R = 
0.00/0.25	  f6(x1) -> f5(x1) 
0.00/0.25	  f5(I0) -> f1(I0) 
0.00/0.25	  f4(I1) -> f3(I1) 
0.00/0.25	  f3(I2) -> f4(I2) 
0.00/0.25	  f2(I3) -> f1(I3) 
0.00/0.25	  f1(I4) -> f2(I4) [I4 <= I4 /\ I4 <= I4]
0.00/0.25	
0.00/0.25	The dependency graph for this problem is:
0.00/0.25	  0 -> 1
0.00/0.25	  1 -> 5
0.00/0.25	  2 -> 3
0.00/0.25	  3 -> 2
0.00/0.25	  4 -> 5
0.00/0.25	  5 -> 4
0.00/0.25	Where:
0.00/0.25	  0) f6#(x1) -> f5#(x1) 
0.00/0.25	  1) f5#(I0) -> f1#(I0) 
0.00/0.25	  2) f4#(I1) -> f3#(I1) 
0.00/0.25	  3) f3#(I2) -> f4#(I2) 
0.00/0.25	  4) f2#(I3) -> f1#(I3) 
0.00/0.25	  5) f1#(I4) -> f2#(I4) [I4 <= I4 /\ I4 <= I4]
0.00/0.25	
0.00/0.25	We have the following SCCs.
0.00/0.25	  { 2, 3 }
0.00/0.25	  { 4, 5 }
0.00/0.25	
0.00/0.25	DP problem for innermost termination.
0.00/0.25	P =
0.00/0.25	  f2#(I3) -> f1#(I3) 
0.00/0.25	  f1#(I4) -> f2#(I4) [I4 <= I4 /\ I4 <= I4]
0.00/0.25	R = 
0.00/0.25	  f6(x1) -> f5(x1) 
0.00/0.25	  f5(I0) -> f1(I0) 
0.00/0.25	  f4(I1) -> f3(I1) 
0.00/0.25	  f3(I2) -> f4(I2) 
0.00/0.25	  f2(I3) -> f1(I3) 
0.00/0.25	  f1(I4) -> f2(I4) [I4 <= I4 /\ I4 <= I4]
0.00/0.25	
0.00/3.23	EOF