2.83/2.79	MAYBE
2.83/2.79	
2.83/2.79	DP problem for innermost termination.
2.83/2.79	P =
2.83/2.79	  f6#(x1, x2, x3) -> f1#(x1, x2, x3) 
2.83/2.79	  f5#(I0, I1, I2) -> f2#(I0, I1, I2) 
2.83/2.79	  f4#(I3, I4, I5) -> f5#(I3, I4 + I5, I5) 
2.83/2.79	  f3#(I6, I7, I8) -> f4#(I6, I7, I8) [I6 = I6]
2.83/2.79	  f2#(I9, I10, I11) -> f3#(I9, I10, I11) [1 <= I10]
2.83/2.79	  f1#(I12, I13, I14) -> f2#(I12, I13, I14) 
2.83/2.79	R = 
2.83/2.79	  f6(x1, x2, x3) -> f1(x1, x2, x3) 
2.83/2.79	  f5(I0, I1, I2) -> f2(I0, I1, I2) 
2.83/2.79	  f4(I3, I4, I5) -> f5(I3, I4 + I5, I5) 
2.83/2.79	  f3(I6, I7, I8) -> f4(I6, I7, I8) [I6 = I6]
2.83/2.79	  f2(I9, I10, I11) -> f3(I9, I10, I11) [1 <= I10]
2.83/2.79	  f1(I12, I13, I14) -> f2(I12, I13, I14) 
2.83/2.79	
2.83/2.79	The dependency graph for this problem is:
2.83/2.79	  0 -> 5
2.83/2.79	  1 -> 4
2.83/2.79	  2 -> 1
2.83/2.79	  3 -> 2
2.83/2.79	  4 -> 3
2.83/2.79	  5 -> 4
2.83/2.79	Where:
2.83/2.79	  0) f6#(x1, x2, x3) -> f1#(x1, x2, x3) 
2.83/2.79	  1) f5#(I0, I1, I2) -> f2#(I0, I1, I2) 
2.83/2.79	  2) f4#(I3, I4, I5) -> f5#(I3, I4 + I5, I5) 
2.83/2.79	  3) f3#(I6, I7, I8) -> f4#(I6, I7, I8) [I6 = I6]
2.83/2.79	  4) f2#(I9, I10, I11) -> f3#(I9, I10, I11) [1 <= I10]
2.83/2.79	  5) f1#(I12, I13, I14) -> f2#(I12, I13, I14) 
2.83/2.79	
2.83/2.79	We have the following SCCs.
2.83/2.79	  { 1, 2, 3, 4 }
2.83/2.79	
2.83/2.79	DP problem for innermost termination.
2.83/2.79	P =
2.83/2.79	  f5#(I0, I1, I2) -> f2#(I0, I1, I2) 
2.83/2.79	  f4#(I3, I4, I5) -> f5#(I3, I4 + I5, I5) 
2.83/2.79	  f3#(I6, I7, I8) -> f4#(I6, I7, I8) [I6 = I6]
2.83/2.79	  f2#(I9, I10, I11) -> f3#(I9, I10, I11) [1 <= I10]
2.83/2.79	R = 
2.83/2.79	  f6(x1, x2, x3) -> f1(x1, x2, x3) 
2.83/2.79	  f5(I0, I1, I2) -> f2(I0, I1, I2) 
2.83/2.79	  f4(I3, I4, I5) -> f5(I3, I4 + I5, I5) 
2.83/2.79	  f3(I6, I7, I8) -> f4(I6, I7, I8) [I6 = I6]
2.83/2.79	  f2(I9, I10, I11) -> f3(I9, I10, I11) [1 <= I10]
2.83/2.79	  f1(I12, I13, I14) -> f2(I12, I13, I14) 
2.83/2.79	
2.83/5.77	EOF