0.00/0.51	MAYBE
0.00/0.51	
0.00/0.51	DP problem for innermost termination.
0.00/0.51	P =
0.00/0.51	  f5#(x1, x2) -> f4#(x1, x2) 
0.00/0.51	  f4#(I0, I1) -> f1#(I0, I1) [I0 <= I1 /\ I1 <= I0]
0.00/0.51	  f3#(I2, I3) -> f1#(I2, I3) 
0.00/0.51	  f1#(I4, I5) -> f3#(1 + I4, 1 + I5) 
0.00/0.51	R = 
0.00/0.51	  f5(x1, x2) -> f4(x1, x2) 
0.00/0.51	  f4(I0, I1) -> f1(I0, I1) [I0 <= I1 /\ I1 <= I0]
0.00/0.51	  f3(I2, I3) -> f1(I2, I3) 
0.00/0.51	  f1(I4, I5) -> f3(1 + I4, 1 + I5) 
0.00/0.51	  f1(I6, I7) -> f2(I6, I7) [1 + I7 <= I6]
0.00/0.51	
0.00/0.51	The dependency graph for this problem is:
0.00/0.51	  0 -> 1
0.00/0.51	  1 -> 3
0.00/0.51	  2 -> 3
0.00/0.51	  3 -> 2
0.00/0.51	Where:
0.00/0.51	  0) f5#(x1, x2) -> f4#(x1, x2) 
0.00/0.51	  1) f4#(I0, I1) -> f1#(I0, I1) [I0 <= I1 /\ I1 <= I0]
0.00/0.51	  2) f3#(I2, I3) -> f1#(I2, I3) 
0.00/0.51	  3) f1#(I4, I5) -> f3#(1 + I4, 1 + I5) 
0.00/0.51	
0.00/0.51	We have the following SCCs.
0.00/0.51	  { 2, 3 }
0.00/0.51	
0.00/0.51	DP problem for innermost termination.
0.00/0.51	P =
0.00/0.51	  f3#(I2, I3) -> f1#(I2, I3) 
0.00/0.51	  f1#(I4, I5) -> f3#(1 + I4, 1 + I5) 
0.00/0.51	R = 
0.00/0.51	  f5(x1, x2) -> f4(x1, x2) 
0.00/0.51	  f4(I0, I1) -> f1(I0, I1) [I0 <= I1 /\ I1 <= I0]
0.00/0.51	  f3(I2, I3) -> f1(I2, I3) 
0.00/0.51	  f1(I4, I5) -> f3(1 + I4, 1 + I5) 
0.00/0.51	  f1(I6, I7) -> f2(I6, I7) [1 + I7 <= I6]
0.00/0.51	
0.00/3.49	EOF