0.00/0.48	MAYBE
0.00/0.48	
0.00/0.48	DP problem for innermost termination.
0.00/0.48	P =
0.00/0.48	  f5#(x1, x2) -> f1#(x1, x2) 
0.00/0.48	  f3#(I2, I3) -> f2#(I2, I3) 
0.00/0.48	  f2#(I4, I5) -> f3#(1 + I4, I5) 
0.00/0.48	  f1#(I6, I7) -> f2#(1, 1) 
0.00/0.48	R = 
0.00/0.48	  f5(x1, x2) -> f1(x1, x2) 
0.00/0.48	  f2(I0, I1) -> f4(I0, I1) [2 <= 0]
0.00/0.48	  f3(I2, I3) -> f2(I2, I3) 
0.00/0.48	  f2(I4, I5) -> f3(1 + I4, I5) 
0.00/0.48	  f1(I6, I7) -> f2(1, 1) 
0.00/0.48	
0.00/0.48	The dependency graph for this problem is:
0.00/0.48	  0 -> 3
0.00/0.48	  1 -> 2
0.00/0.48	  2 -> 1
0.00/0.48	  3 -> 2
0.00/0.48	Where:
0.00/0.48	  0) f5#(x1, x2) -> f1#(x1, x2) 
0.00/0.48	  1) f3#(I2, I3) -> f2#(I2, I3) 
0.00/0.48	  2) f2#(I4, I5) -> f3#(1 + I4, I5) 
0.00/0.48	  3) f1#(I6, I7) -> f2#(1, 1) 
0.00/0.48	
0.00/0.48	We have the following SCCs.
0.00/0.48	  { 1, 2 }
0.00/0.48	
0.00/0.48	DP problem for innermost termination.
0.00/0.48	P =
0.00/0.48	  f3#(I2, I3) -> f2#(I2, I3) 
0.00/0.48	  f2#(I4, I5) -> f3#(1 + I4, I5) 
0.00/0.48	R = 
0.00/0.48	  f5(x1, x2) -> f1(x1, x2) 
0.00/0.48	  f2(I0, I1) -> f4(I0, I1) [2 <= 0]
0.00/0.48	  f3(I2, I3) -> f2(I2, I3) 
0.00/0.48	  f2(I4, I5) -> f3(1 + I4, I5) 
0.00/0.48	  f1(I6, I7) -> f2(1, 1) 
0.00/0.48	
0.00/3.46	EOF