0.00/0.15	MAYBE
0.00/0.15	
0.00/0.15	DP problem for innermost termination.
0.00/0.15	P =
0.00/0.15	  f4#(x1) -> f3#(x1) 
0.00/0.15	  f3#(I0) -> f1#(0) [y1 = 0 /\ y2 = 1]
0.00/0.15	  f2#(I1) -> f1#(I1) 
0.00/0.15	  f1#(I2) -> f2#(I2) 
0.00/0.15	R = 
0.00/0.15	  f4(x1) -> f3(x1) 
0.00/0.15	  f3(I0) -> f1(0) [y1 = 0 /\ y2 = 1]
0.00/0.15	  f2(I1) -> f1(I1) 
0.00/0.15	  f1(I2) -> f2(I2) 
0.00/0.15	
0.00/0.15	The dependency graph for this problem is:
0.00/0.15	  0 -> 1
0.00/0.15	  1 -> 3
0.00/0.15	  2 -> 3
0.00/0.15	  3 -> 2
0.00/0.15	Where:
0.00/0.15	  0) f4#(x1) -> f3#(x1) 
0.00/0.15	  1) f3#(I0) -> f1#(0) [y1 = 0 /\ y2 = 1]
0.00/0.15	  2) f2#(I1) -> f1#(I1) 
0.00/0.15	  3) f1#(I2) -> f2#(I2) 
0.00/0.15	
0.00/0.15	We have the following SCCs.
0.00/0.15	  { 2, 3 }
0.00/0.15	
0.00/0.15	DP problem for innermost termination.
0.00/0.15	P =
0.00/0.15	  f2#(I1) -> f1#(I1) 
0.00/0.15	  f1#(I2) -> f2#(I2) 
0.00/0.15	R = 
0.00/0.15	  f4(x1) -> f3(x1) 
0.00/0.15	  f3(I0) -> f1(0) [y1 = 0 /\ y2 = 1]
0.00/0.15	  f2(I1) -> f1(I1) 
0.00/0.15	  f1(I2) -> f2(I2) 
0.00/0.15	
0.00/3.13	EOF