0.93/0.98	MAYBE
0.93/0.98	
0.93/0.98	DP problem for innermost termination.
0.93/0.98	P =
0.93/0.98	  f4#(x1, x2) -> f3#(x1, x2) 
0.93/0.98	  f3#(I0, I1) -> f1#(I0, I1) [1 <= I1]
0.93/0.98	  f2#(I2, I3) -> f1#(I2, I3) 
0.93/0.98	  f1#(I4, I5) -> f2#(I4 - I5, 1 + I5) [1 <= I4]
0.93/0.98	R = 
0.93/0.98	  f4(x1, x2) -> f3(x1, x2) 
0.93/0.98	  f3(I0, I1) -> f1(I0, I1) [1 <= I1]
0.93/0.98	  f2(I2, I3) -> f1(I2, I3) 
0.93/0.98	  f1(I4, I5) -> f2(I4 - I5, 1 + I5) [1 <= I4]
0.93/0.98	
0.93/0.98	The dependency graph for this problem is:
0.93/0.98	  0 -> 1
0.93/0.98	  1 -> 3
0.93/0.98	  2 -> 3
0.93/0.98	  3 -> 2
0.93/0.98	Where:
0.93/0.98	  0) f4#(x1, x2) -> f3#(x1, x2) 
0.93/0.98	  1) f3#(I0, I1) -> f1#(I0, I1) [1 <= I1]
0.93/0.98	  2) f2#(I2, I3) -> f1#(I2, I3) 
0.93/0.98	  3) f1#(I4, I5) -> f2#(I4 - I5, 1 + I5) [1 <= I4]
0.93/0.98	
0.93/0.98	We have the following SCCs.
0.93/0.98	  { 2, 3 }
0.93/0.98	
0.93/0.98	DP problem for innermost termination.
0.93/0.98	P =
0.93/0.98	  f2#(I2, I3) -> f1#(I2, I3) 
0.93/0.98	  f1#(I4, I5) -> f2#(I4 - I5, 1 + I5) [1 <= I4]
0.93/0.98	R = 
0.93/0.98	  f4(x1, x2) -> f3(x1, x2) 
0.93/0.98	  f3(I0, I1) -> f1(I0, I1) [1 <= I1]
0.93/0.98	  f2(I2, I3) -> f1(I2, I3) 
0.93/0.98	  f1(I4, I5) -> f2(I4 - I5, 1 + I5) [1 <= I4]
0.93/0.98	
0.93/3.96	EOF