0.79/0.82	MAYBE
0.79/0.82	
0.79/0.82	DP problem for innermost termination.
0.79/0.82	P =
0.79/0.82	  f5#(x1, x2) -> f4#(x1, x2) 
0.79/0.82	  f4#(I0, I1) -> f1#(I0, I1) [1 <= I1]
0.79/0.82	  f3#(I2, I3) -> f1#(I2, I3) 
0.79/0.82	  f2#(I4, I5) -> f3#(I4, I5) [I4 = I4]
0.79/0.82	  f1#(I6, I7) -> f2#(I6, -1 + I7) 
0.79/0.82	R = 
0.79/0.82	  f5(x1, x2) -> f4(x1, x2) 
0.79/0.82	  f4(I0, I1) -> f1(I0, I1) [1 <= I1]
0.79/0.82	  f3(I2, I3) -> f1(I2, I3) 
0.79/0.82	  f2(I4, I5) -> f3(I4, I5) [I4 = I4]
0.79/0.82	  f1(I6, I7) -> f2(I6, -1 + I7) 
0.79/0.82	
0.79/0.82	The dependency graph for this problem is:
0.79/0.82	  0 -> 1
0.79/0.82	  1 -> 4
0.79/0.82	  2 -> 4
0.79/0.82	  3 -> 2
0.79/0.82	  4 -> 3
0.79/0.82	Where:
0.79/0.82	  0) f5#(x1, x2) -> f4#(x1, x2) 
0.79/0.82	  1) f4#(I0, I1) -> f1#(I0, I1) [1 <= I1]
0.79/0.82	  2) f3#(I2, I3) -> f1#(I2, I3) 
0.79/0.82	  3) f2#(I4, I5) -> f3#(I4, I5) [I4 = I4]
0.79/0.82	  4) f1#(I6, I7) -> f2#(I6, -1 + I7) 
0.79/0.82	
0.79/0.82	We have the following SCCs.
0.79/0.82	  { 2, 3, 4 }
0.79/0.82	
0.79/0.82	DP problem for innermost termination.
0.79/0.82	P =
0.79/0.82	  f3#(I2, I3) -> f1#(I2, I3) 
0.79/0.82	  f2#(I4, I5) -> f3#(I4, I5) [I4 = I4]
0.79/0.82	  f1#(I6, I7) -> f2#(I6, -1 + I7) 
0.79/0.82	R = 
0.79/0.82	  f5(x1, x2) -> f4(x1, x2) 
0.79/0.82	  f4(I0, I1) -> f1(I0, I1) [1 <= I1]
0.79/0.82	  f3(I2, I3) -> f1(I2, I3) 
0.79/0.82	  f2(I4, I5) -> f3(I4, I5) [I4 = I4]
0.79/0.82	  f1(I6, I7) -> f2(I6, -1 + I7) 
0.79/0.82	
0.79/3.80	EOF