0.87/0.93	MAYBE
0.87/0.93	
0.87/0.93	DP problem for innermost termination.
0.87/0.93	P =
0.87/0.93	  f4#(x1, x2) -> f3#(x1, x2) 
0.87/0.93	  f3#(I0, I1) -> f1#(I0, 1) 
0.87/0.93	  f2#(I2, I3) -> f1#(I2, I3) 
0.87/0.93	  f1#(I4, I5) -> f2#(I4 - 3 * I5, I5) [0 <= I4]
0.87/0.93	R = 
0.87/0.93	  f4(x1, x2) -> f3(x1, x2) 
0.87/0.93	  f3(I0, I1) -> f1(I0, 1) 
0.87/0.93	  f2(I2, I3) -> f1(I2, I3) 
0.87/0.93	  f1(I4, I5) -> f2(I4 - 3 * I5, I5) [0 <= I4]
0.87/0.93	
0.87/0.93	The dependency graph for this problem is:
0.87/0.93	  0 -> 1
0.87/0.93	  1 -> 3
0.87/0.93	  2 -> 3
0.87/0.93	  3 -> 2
0.87/0.93	Where:
0.87/0.93	  0) f4#(x1, x2) -> f3#(x1, x2) 
0.87/0.93	  1) f3#(I0, I1) -> f1#(I0, 1) 
0.87/0.93	  2) f2#(I2, I3) -> f1#(I2, I3) 
0.87/0.93	  3) f1#(I4, I5) -> f2#(I4 - 3 * I5, I5) [0 <= I4]
0.87/0.93	
0.87/0.93	We have the following SCCs.
0.87/0.93	  { 2, 3 }
0.87/0.93	
0.87/0.93	DP problem for innermost termination.
0.87/0.93	P =
0.87/0.93	  f2#(I2, I3) -> f1#(I2, I3) 
0.87/0.93	  f1#(I4, I5) -> f2#(I4 - 3 * I5, I5) [0 <= I4]
0.87/0.93	R = 
0.87/0.93	  f4(x1, x2) -> f3(x1, x2) 
0.87/0.93	  f3(I0, I1) -> f1(I0, 1) 
0.87/0.93	  f2(I2, I3) -> f1(I2, I3) 
0.87/0.93	  f1(I4, I5) -> f2(I4 - 3 * I5, I5) [0 <= I4]
0.87/0.93	
0.87/3.91	EOF