2.21/2.30	MAYBE
2.21/2.30	
2.21/2.30	DP problem for innermost termination.
2.21/2.30	P =
2.21/2.30	  f4#(x1, x2) -> f3#(x1, x2) 
2.21/2.30	  f3#(I0, I1) -> f2#(I0, I1) [1 <= I1 /\ 1 <= I0]
2.21/2.30	  f2#(I2, I3) -> f1#(-1 + I2, I3) 
2.21/2.30	  f2#(I4, I5) -> f1#(I4, -1 + I5) 
2.21/2.30	  f1#(I6, I7) -> f2#(I6, I7) [1 <= I7 /\ 1 <= I6]
2.21/2.30	R = 
2.21/2.30	  f4(x1, x2) -> f3(x1, x2) 
2.21/2.30	  f3(I0, I1) -> f2(I0, I1) [1 <= I1 /\ 1 <= I0]
2.21/2.30	  f2(I2, I3) -> f1(-1 + I2, I3) 
2.21/2.30	  f2(I4, I5) -> f1(I4, -1 + I5) 
2.21/2.30	  f1(I6, I7) -> f2(I6, I7) [1 <= I7 /\ 1 <= I6]
2.21/2.30	
2.21/2.30	The dependency graph for this problem is:
2.21/2.30	  0 -> 1
2.21/2.30	  1 -> 2, 3
2.21/2.30	  2 -> 4
2.21/2.30	  3 -> 4
2.21/2.30	  4 -> 2, 3
2.21/2.30	Where:
2.21/2.30	  0) f4#(x1, x2) -> f3#(x1, x2) 
2.21/2.30	  1) f3#(I0, I1) -> f2#(I0, I1) [1 <= I1 /\ 1 <= I0]
2.21/2.30	  2) f2#(I2, I3) -> f1#(-1 + I2, I3) 
2.21/2.30	  3) f2#(I4, I5) -> f1#(I4, -1 + I5) 
2.21/2.30	  4) f1#(I6, I7) -> f2#(I6, I7) [1 <= I7 /\ 1 <= I6]
2.21/2.30	
2.21/2.30	We have the following SCCs.
2.21/2.30	  { 2, 3, 4 }
2.21/2.30	
2.21/2.30	DP problem for innermost termination.
2.21/2.30	P =
2.21/2.30	  f2#(I2, I3) -> f1#(-1 + I2, I3) 
2.21/2.30	  f2#(I4, I5) -> f1#(I4, -1 + I5) 
2.21/2.30	  f1#(I6, I7) -> f2#(I6, I7) [1 <= I7 /\ 1 <= I6]
2.21/2.30	R = 
2.21/2.30	  f4(x1, x2) -> f3(x1, x2) 
2.21/2.30	  f3(I0, I1) -> f2(I0, I1) [1 <= I1 /\ 1 <= I0]
2.21/2.30	  f2(I2, I3) -> f1(-1 + I2, I3) 
2.21/2.30	  f2(I4, I5) -> f1(I4, -1 + I5) 
2.21/2.30	  f1(I6, I7) -> f2(I6, I7) [1 <= I7 /\ 1 <= I6]
2.21/2.30	
2.21/2.30	EOF