2.23/2.26	MAYBE
2.23/2.26	
2.23/2.26	DP problem for innermost termination.
2.23/2.26	P =
2.23/2.26	  init#(x1, x2) -> f1#(rnd1, rnd2) 
2.23/2.26	  f3#(I0, I1) -> f2#(I0 + 1, I2) [I1 <= 1 /\ -1 <= I0 - 1]
2.23/2.26	  f3#(I3, I4) -> f3#(I3 + 1, I4 - 2) [1 <= I4 - 1]
2.23/2.26	  f2#(I5, I6) -> f3#(0, I5 - 2) [1 <= I5 - 1]
2.23/2.26	  f1#(I7, I8) -> f2#(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
2.23/2.26	R = 
2.23/2.26	  init(x1, x2) -> f1(rnd1, rnd2) 
2.23/2.26	  f3(I0, I1) -> f2(I0 + 1, I2) [I1 <= 1 /\ -1 <= I0 - 1]
2.23/2.26	  f3(I3, I4) -> f3(I3 + 1, I4 - 2) [1 <= I4 - 1]
2.23/2.26	  f2(I5, I6) -> f3(0, I5 - 2) [1 <= I5 - 1]
2.23/2.26	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
2.23/2.26	
2.23/2.26	The dependency graph for this problem is:
2.23/2.26	  0 -> 4
2.23/2.26	  1 -> 3
2.23/2.26	  2 -> 1, 2
2.23/2.26	  3 -> 1, 2
2.23/2.26	  4 -> 3
2.23/2.26	Where:
2.23/2.26	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
2.23/2.26	  1) f3#(I0, I1) -> f2#(I0 + 1, I2) [I1 <= 1 /\ -1 <= I0 - 1]
2.23/2.26	  2) f3#(I3, I4) -> f3#(I3 + 1, I4 - 2) [1 <= I4 - 1]
2.23/2.26	  3) f2#(I5, I6) -> f3#(0, I5 - 2) [1 <= I5 - 1]
2.23/2.26	  4) f1#(I7, I8) -> f2#(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
2.23/2.26	
2.23/2.26	We have the following SCCs.
2.23/2.26	  { 1, 2, 3 }
2.23/2.26	
2.23/2.26	DP problem for innermost termination.
2.23/2.26	P =
2.23/2.26	  f3#(I0, I1) -> f2#(I0 + 1, I2) [I1 <= 1 /\ -1 <= I0 - 1]
2.23/2.26	  f3#(I3, I4) -> f3#(I3 + 1, I4 - 2) [1 <= I4 - 1]
2.23/2.26	  f2#(I5, I6) -> f3#(0, I5 - 2) [1 <= I5 - 1]
2.23/2.26	R = 
2.23/2.26	  init(x1, x2) -> f1(rnd1, rnd2) 
2.23/2.26	  f3(I0, I1) -> f2(I0 + 1, I2) [I1 <= 1 /\ -1 <= I0 - 1]
2.23/2.26	  f3(I3, I4) -> f3(I3 + 1, I4 - 2) [1 <= I4 - 1]
2.23/2.26	  f2(I5, I6) -> f3(0, I5 - 2) [1 <= I5 - 1]
2.23/2.26	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
2.23/2.26	
2.23/2.26	EOF