0.00/0.21	MAYBE
0.00/0.21	
0.00/0.21	DP problem for innermost termination.
0.00/0.21	P =
0.00/0.21	  f5#(x1) -> f4#(x1) 
0.00/0.21	  f4#(I0) -> f3#(1) [y1 = 0]
0.00/0.21	  f3#(I1) -> f1#(0) 
0.00/0.21	  f3#(I2) -> f1#(1) 
0.00/0.21	  f2#(I3) -> f1#(I3) 
0.00/0.21	  f1#(I4) -> f2#(I4) 
0.00/0.21	R = 
0.00/0.21	  f5(x1) -> f4(x1) 
0.00/0.21	  f4(I0) -> f3(1) [y1 = 0]
0.00/0.21	  f3(I1) -> f1(0) 
0.00/0.21	  f3(I2) -> f1(1) 
0.00/0.21	  f2(I3) -> f1(I3) 
0.00/0.21	  f1(I4) -> f2(I4) 
0.00/0.21	
0.00/0.21	The dependency graph for this problem is:
0.00/0.21	  0 -> 1
0.00/0.21	  1 -> 2, 3
0.00/0.21	  2 -> 5
0.00/0.21	  3 -> 5
0.00/0.21	  4 -> 5
0.00/0.21	  5 -> 4
0.00/0.21	Where:
0.00/0.21	  0) f5#(x1) -> f4#(x1) 
0.00/0.21	  1) f4#(I0) -> f3#(1) [y1 = 0]
0.00/0.21	  2) f3#(I1) -> f1#(0) 
0.00/0.21	  3) f3#(I2) -> f1#(1) 
0.00/0.21	  4) f2#(I3) -> f1#(I3) 
0.00/0.21	  5) f1#(I4) -> f2#(I4) 
0.00/0.21	
0.00/0.21	We have the following SCCs.
0.00/0.21	  { 4, 5 }
0.00/0.21	
0.00/0.21	DP problem for innermost termination.
0.00/0.21	P =
0.00/0.21	  f2#(I3) -> f1#(I3) 
0.00/0.21	  f1#(I4) -> f2#(I4) 
0.00/0.21	R = 
0.00/0.21	  f5(x1) -> f4(x1) 
0.00/0.21	  f4(I0) -> f3(1) [y1 = 0]
0.00/0.21	  f3(I1) -> f1(0) 
0.00/0.21	  f3(I2) -> f1(1) 
0.00/0.21	  f2(I3) -> f1(I3) 
0.00/0.21	  f1(I4) -> f2(I4) 
0.00/0.21	
0.00/3.19	EOF