0.61/0.67	MAYBE
0.61/0.67	
0.61/0.67	DP problem for innermost termination.
0.61/0.67	P =
0.61/0.67	  f6#(x1, x2) -> f3#(x1, x2) 
0.61/0.67	  f4#(I0, I1) -> f1#(-1, I1) [I1 <= 0]
0.61/0.67	  f5#(I2, I3) -> f4#(I2, I3) 
0.61/0.67	  f4#(I4, I5) -> f5#(I4, -1 + I5) [1 <= I5]
0.61/0.67	  f3#(I6, I7) -> f4#(0, I7) 
0.61/0.67	  f2#(I8, I9) -> f1#(I8, I9) 
0.61/0.67	  f1#(I10, I11) -> f2#(I10, I11) 
0.61/0.67	R = 
0.61/0.67	  f6(x1, x2) -> f3(x1, x2) 
0.61/0.67	  f4(I0, I1) -> f1(-1, I1) [I1 <= 0]
0.61/0.67	  f5(I2, I3) -> f4(I2, I3) 
0.61/0.67	  f4(I4, I5) -> f5(I4, -1 + I5) [1 <= I5]
0.61/0.67	  f3(I6, I7) -> f4(0, I7) 
0.61/0.67	  f2(I8, I9) -> f1(I8, I9) 
0.61/0.67	  f1(I10, I11) -> f2(I10, I11) 
0.61/0.67	
0.61/0.67	The dependency graph for this problem is:
0.61/0.67	  0 -> 4
0.61/0.67	  1 -> 6
0.61/0.67	  2 -> 1, 3
0.61/0.67	  3 -> 2
0.61/0.67	  4 -> 1, 3
0.61/0.67	  5 -> 6
0.61/0.67	  6 -> 5
0.61/0.67	Where:
0.61/0.67	  0) f6#(x1, x2) -> f3#(x1, x2) 
0.61/0.67	  1) f4#(I0, I1) -> f1#(-1, I1) [I1 <= 0]
0.61/0.67	  2) f5#(I2, I3) -> f4#(I2, I3) 
0.61/0.67	  3) f4#(I4, I5) -> f5#(I4, -1 + I5) [1 <= I5]
0.61/0.67	  4) f3#(I6, I7) -> f4#(0, I7) 
0.61/0.67	  5) f2#(I8, I9) -> f1#(I8, I9) 
0.61/0.67	  6) f1#(I10, I11) -> f2#(I10, I11) 
0.61/0.67	
0.61/0.67	We have the following SCCs.
0.61/0.67	  { 2, 3 }
0.61/0.67	  { 5, 6 }
0.61/0.67	
0.61/0.67	DP problem for innermost termination.
0.61/0.67	P =
0.61/0.67	  f2#(I8, I9) -> f1#(I8, I9) 
0.61/0.67	  f1#(I10, I11) -> f2#(I10, I11) 
0.61/0.67	R = 
0.61/0.67	  f6(x1, x2) -> f3(x1, x2) 
0.61/0.67	  f4(I0, I1) -> f1(-1, I1) [I1 <= 0]
0.61/0.67	  f5(I2, I3) -> f4(I2, I3) 
0.61/0.67	  f4(I4, I5) -> f5(I4, -1 + I5) [1 <= I5]
0.61/0.67	  f3(I6, I7) -> f4(0, I7) 
0.61/0.67	  f2(I8, I9) -> f1(I8, I9) 
0.61/0.67	  f1(I10, I11) -> f2(I10, I11) 
0.61/0.67	
0.61/3.65	EOF