0.00/0.24	MAYBE
0.00/0.24	
0.00/0.24	DP problem for innermost termination.
0.00/0.24	P =
0.00/0.24	  f6#(x1) -> f5#(x1) 
0.00/0.24	  f5#(I0) -> f3#(2) 
0.00/0.24	  f4#(I1) -> f3#(I1) 
0.00/0.24	  f3#(I2) -> f4#(2) [y1 = 3]
0.00/0.24	  f2#(I3) -> f1#(I3) 
0.00/0.24	  f1#(I4) -> f2#(2) 
0.00/0.24	R = 
0.00/0.24	  f6(x1) -> f5(x1) 
0.00/0.24	  f5(I0) -> f3(2) 
0.00/0.24	  f4(I1) -> f3(I1) 
0.00/0.24	  f3(I2) -> f4(2) [y1 = 3]
0.00/0.24	  f2(I3) -> f1(I3) 
0.00/0.24	  f1(I4) -> f2(2) 
0.00/0.24	
0.00/0.24	The dependency graph for this problem is:
0.00/0.24	  0 -> 1
0.00/0.24	  1 -> 3
0.00/0.24	  2 -> 3
0.00/0.24	  3 -> 2
0.00/0.24	  4 -> 5
0.00/0.24	  5 -> 4
0.00/0.24	Where:
0.00/0.24	  0) f6#(x1) -> f5#(x1) 
0.00/0.24	  1) f5#(I0) -> f3#(2) 
0.00/0.24	  2) f4#(I1) -> f3#(I1) 
0.00/0.24	  3) f3#(I2) -> f4#(2) [y1 = 3]
0.00/0.24	  4) f2#(I3) -> f1#(I3) 
0.00/0.24	  5) f1#(I4) -> f2#(2) 
0.00/0.24	
0.00/0.24	We have the following SCCs.
0.00/0.24	  { 2, 3 }
0.00/0.24	  { 4, 5 }
0.00/0.24	
0.00/0.24	DP problem for innermost termination.
0.00/0.24	P =
0.00/0.24	  f2#(I3) -> f1#(I3) 
0.00/0.24	  f1#(I4) -> f2#(2) 
0.00/0.24	R = 
0.00/0.24	  f6(x1) -> f5(x1) 
0.00/0.24	  f5(I0) -> f3(2) 
0.00/0.24	  f4(I1) -> f3(I1) 
0.00/0.24	  f3(I2) -> f4(2) [y1 = 3]
0.00/0.24	  f2(I3) -> f1(I3) 
0.00/0.24	  f1(I4) -> f2(2) 
0.00/0.24	
0.00/3.22	EOF