0.00/0.49	MAYBE
0.00/0.49	
0.00/0.49	DP problem for innermost termination.
0.00/0.49	P =
0.00/0.49	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.49	  f3#(I0, I1, I2) -> f3#(I0, I1 + 1, I1 + 1) [I1 = I2 /\ 0 <= I1 - 1]
0.00/0.49	  f3#(I3, I4, I5) -> f2#(I3 + 1, I6, I7) [0 = I5 /\ 0 = I4]
0.00/0.49	  f2#(I8, I9, I10) -> f3#(I8, I8, I8) [I8 <= 9]
0.00/0.49	  f1#(I11, I12, I13) -> f2#(I12, I14, I15) [-1 <= I12 - 1 /\ 0 <= I11 - 1]
0.00/0.49	R = 
0.00/0.49	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.49	  f3(I0, I1, I2) -> f3(I0, I1 + 1, I1 + 1) [I1 = I2 /\ 0 <= I1 - 1]
0.00/0.49	  f3(I3, I4, I5) -> f2(I3 + 1, I6, I7) [0 = I5 /\ 0 = I4]
0.00/0.49	  f2(I8, I9, I10) -> f3(I8, I8, I8) [I8 <= 9]
0.00/0.49	  f1(I11, I12, I13) -> f2(I12, I14, I15) [-1 <= I12 - 1 /\ 0 <= I11 - 1]
0.00/0.49	
0.00/0.49	The dependency graph for this problem is:
0.00/0.49	  0 -> 4
0.00/0.49	  1 -> 1
0.00/0.49	  2 -> 3
0.00/0.49	  3 -> 1, 2
0.00/0.49	  4 -> 3
0.00/0.49	Where:
0.00/0.49	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.49	  1) f3#(I0, I1, I2) -> f3#(I0, I1 + 1, I1 + 1) [I1 = I2 /\ 0 <= I1 - 1]
0.00/0.49	  2) f3#(I3, I4, I5) -> f2#(I3 + 1, I6, I7) [0 = I5 /\ 0 = I4]
0.00/0.49	  3) f2#(I8, I9, I10) -> f3#(I8, I8, I8) [I8 <= 9]
0.00/0.49	  4) f1#(I11, I12, I13) -> f2#(I12, I14, I15) [-1 <= I12 - 1 /\ 0 <= I11 - 1]
0.00/0.49	
0.00/0.49	We have the following SCCs.
0.00/0.49	  { 2, 3 }
0.00/0.49	  { 1 }
0.00/0.49	
0.00/0.49	DP problem for innermost termination.
0.00/0.49	P =
0.00/0.49	  f3#(I0, I1, I2) -> f3#(I0, I1 + 1, I1 + 1) [I1 = I2 /\ 0 <= I1 - 1]
0.00/0.49	R = 
0.00/0.49	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.49	  f3(I0, I1, I2) -> f3(I0, I1 + 1, I1 + 1) [I1 = I2 /\ 0 <= I1 - 1]
0.00/0.49	  f3(I3, I4, I5) -> f2(I3 + 1, I6, I7) [0 = I5 /\ 0 = I4]
0.00/0.49	  f2(I8, I9, I10) -> f3(I8, I8, I8) [I8 <= 9]
0.00/0.49	  f1(I11, I12, I13) -> f2(I12, I14, I15) [-1 <= I12 - 1 /\ 0 <= I11 - 1]
0.00/0.49	
0.00/3.47	EOF