2.72/2.71	MAYBE
2.72/2.71	
2.72/2.71	DP problem for innermost termination.
2.72/2.71	P =
2.72/2.71	  f4#(x1, x2, x3) -> f3#(x1, x2, x3) 
2.72/2.71	  f3#(I0, I1, I2) -> f1#(I0, I1, I2) 
2.72/2.71	  f2#(I3, I4, I5) -> f1#(-1 + I3, I4, I5) 
2.72/2.71	  f2#(I6, I7, I8) -> f1#(I8, -1 + I7, I8) 
2.72/2.71	  f1#(I9, I10, I11) -> f2#(I9, I10, I11) [1 <= I9]
2.72/2.71	R = 
2.72/2.71	  f4(x1, x2, x3) -> f3(x1, x2, x3) 
2.72/2.71	  f3(I0, I1, I2) -> f1(I0, I1, I2) 
2.72/2.71	  f2(I3, I4, I5) -> f1(-1 + I3, I4, I5) 
2.72/2.71	  f2(I6, I7, I8) -> f1(I8, -1 + I7, I8) 
2.72/2.71	  f1(I9, I10, I11) -> f2(I9, I10, I11) [1 <= I9]
2.72/2.71	
2.72/2.71	The dependency graph for this problem is:
2.72/2.71	  0 -> 1
2.72/2.71	  1 -> 4
2.72/2.71	  2 -> 4
2.72/2.71	  3 -> 4
2.72/2.71	  4 -> 2, 3
2.72/2.71	Where:
2.72/2.71	  0) f4#(x1, x2, x3) -> f3#(x1, x2, x3) 
2.72/2.71	  1) f3#(I0, I1, I2) -> f1#(I0, I1, I2) 
2.72/2.71	  2) f2#(I3, I4, I5) -> f1#(-1 + I3, I4, I5) 
2.72/2.71	  3) f2#(I6, I7, I8) -> f1#(I8, -1 + I7, I8) 
2.72/2.71	  4) f1#(I9, I10, I11) -> f2#(I9, I10, I11) [1 <= I9]
2.72/2.71	
2.72/2.71	We have the following SCCs.
2.72/2.71	  { 2, 3, 4 }
2.72/2.71	
2.72/2.71	DP problem for innermost termination.
2.72/2.71	P =
2.72/2.71	  f2#(I3, I4, I5) -> f1#(-1 + I3, I4, I5) 
2.72/2.71	  f2#(I6, I7, I8) -> f1#(I8, -1 + I7, I8) 
2.72/2.71	  f1#(I9, I10, I11) -> f2#(I9, I10, I11) [1 <= I9]
2.72/2.71	R = 
2.72/2.71	  f4(x1, x2, x3) -> f3(x1, x2, x3) 
2.72/2.71	  f3(I0, I1, I2) -> f1(I0, I1, I2) 
2.72/2.71	  f2(I3, I4, I5) -> f1(-1 + I3, I4, I5) 
2.72/2.71	  f2(I6, I7, I8) -> f1(I8, -1 + I7, I8) 
2.72/2.71	  f1(I9, I10, I11) -> f2(I9, I10, I11) [1 <= I9]
2.72/2.71	
2.72/2.72	EOF