2.33/2.63	MAYBE
2.33/2.63	
2.33/2.63	DP problem for innermost termination.
2.33/2.63	P =
2.33/2.63	  f6#(x1, x2) -> f1#(x1, x2) 
2.33/2.63	  f5#(I0, I1) -> f2#(I0, I1) 
2.33/2.63	  f4#(I2, I3) -> f5#(I2 + I3, I3) 
2.33/2.63	  f3#(I4, I5) -> f4#(I4, I5) [1 <= I5]
2.33/2.63	  f3#(I6, I7) -> f4#(I6, I7) [1 + I7 <= 0]
2.33/2.63	  f2#(I8, I9) -> f3#(I8, I9) [1 <= I8]
2.33/2.63	  f1#(I10, I11) -> f2#(I10, I11) 
2.33/2.63	R = 
2.33/2.63	  f6(x1, x2) -> f1(x1, x2) 
2.33/2.63	  f5(I0, I1) -> f2(I0, I1) 
2.33/2.63	  f4(I2, I3) -> f5(I2 + I3, I3) 
2.33/2.63	  f3(I4, I5) -> f4(I4, I5) [1 <= I5]
2.33/2.63	  f3(I6, I7) -> f4(I6, I7) [1 + I7 <= 0]
2.33/2.63	  f2(I8, I9) -> f3(I8, I9) [1 <= I8]
2.33/2.63	  f1(I10, I11) -> f2(I10, I11) 
2.33/2.63	
2.33/2.63	The dependency graph for this problem is:
2.33/2.63	  0 -> 6
2.33/2.63	  1 -> 5
2.33/2.63	  2 -> 1
2.33/2.63	  3 -> 2
2.33/2.63	  4 -> 2
2.33/2.63	  5 -> 3, 4
2.33/2.63	  6 -> 5
2.33/2.63	Where:
2.33/2.63	  0) f6#(x1, x2) -> f1#(x1, x2) 
2.33/2.63	  1) f5#(I0, I1) -> f2#(I0, I1) 
2.33/2.63	  2) f4#(I2, I3) -> f5#(I2 + I3, I3) 
2.33/2.63	  3) f3#(I4, I5) -> f4#(I4, I5) [1 <= I5]
2.33/2.63	  4) f3#(I6, I7) -> f4#(I6, I7) [1 + I7 <= 0]
2.33/2.63	  5) f2#(I8, I9) -> f3#(I8, I9) [1 <= I8]
2.33/2.63	  6) f1#(I10, I11) -> f2#(I10, I11) 
2.33/2.63	
2.33/2.63	We have the following SCCs.
2.33/2.63	  { 1, 2, 3, 4, 5 }
2.33/2.63	
2.33/2.63	DP problem for innermost termination.
2.33/2.63	P =
2.33/2.63	  f5#(I0, I1) -> f2#(I0, I1) 
2.33/2.63	  f4#(I2, I3) -> f5#(I2 + I3, I3) 
2.33/2.63	  f3#(I4, I5) -> f4#(I4, I5) [1 <= I5]
2.33/2.63	  f3#(I6, I7) -> f4#(I6, I7) [1 + I7 <= 0]
2.33/2.63	  f2#(I8, I9) -> f3#(I8, I9) [1 <= I8]
2.33/2.63	R = 
2.33/2.63	  f6(x1, x2) -> f1(x1, x2) 
2.33/2.63	  f5(I0, I1) -> f2(I0, I1) 
2.33/2.63	  f4(I2, I3) -> f5(I2 + I3, I3) 
2.33/2.63	  f3(I4, I5) -> f4(I4, I5) [1 <= I5]
2.33/2.63	  f3(I6, I7) -> f4(I6, I7) [1 + I7 <= 0]
2.33/2.63	  f2(I8, I9) -> f3(I8, I9) [1 <= I8]
2.33/2.63	  f1(I10, I11) -> f2(I10, I11) 
2.33/2.63	
2.33/5.61	EOF