1.59/1.69	MAYBE
1.59/1.69	
1.59/1.69	DP problem for innermost termination.
1.59/1.69	P =
1.59/1.69	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
1.59/1.69	  f2#(I0, I1, I2) -> f2#(1, 1, 1) [0 = I2 /\ I0 <= 1]
1.59/1.69	  f2#(I3, I4, I5) -> f2#(1, 1, I5 + 1) [1 = I4 /\ I3 <= 1 /\ I5 <= 9 /\ 0 <= I5 - 1 /\ I5 <= 10]
1.59/1.69	  f2#(I6, I7, I8) -> f2#(0, 0, 9) [10 = I8 /\ I6 <= 1]
1.59/1.69	  f2#(I9, I10, I11) -> f2#(0, 0, I11 - 1) [0 = I10 /\ 0 <= I11 - 1 /\ I9 <= 1 /\ I11 <= 10 /\ I11 <= 9]
1.59/1.69	  f1#(I12, I13, I14) -> f2#(0, 0, I13) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
1.59/1.69	R = 
1.59/1.69	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
1.59/1.69	  f2(I0, I1, I2) -> f2(1, 1, 1) [0 = I2 /\ I0 <= 1]
1.59/1.69	  f2(I3, I4, I5) -> f2(1, 1, I5 + 1) [1 = I4 /\ I3 <= 1 /\ I5 <= 9 /\ 0 <= I5 - 1 /\ I5 <= 10]
1.59/1.69	  f2(I6, I7, I8) -> f2(0, 0, 9) [10 = I8 /\ I6 <= 1]
1.59/1.69	  f2(I9, I10, I11) -> f2(0, 0, I11 - 1) [0 = I10 /\ 0 <= I11 - 1 /\ I9 <= 1 /\ I11 <= 10 /\ I11 <= 9]
1.59/1.69	  f1(I12, I13, I14) -> f2(0, 0, I13) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
1.59/1.69	
1.59/1.69	The dependency graph for this problem is:
1.59/1.69	  0 -> 5
1.59/1.69	  1 -> 2
1.59/1.69	  2 -> 2, 3
1.59/1.69	  3 -> 4
1.59/1.69	  4 -> 1, 4
1.59/1.69	  5 -> 1, 3, 4
1.59/1.69	Where:
1.59/1.69	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
1.59/1.69	  1) f2#(I0, I1, I2) -> f2#(1, 1, 1) [0 = I2 /\ I0 <= 1]
1.59/1.69	  2) f2#(I3, I4, I5) -> f2#(1, 1, I5 + 1) [1 = I4 /\ I3 <= 1 /\ I5 <= 9 /\ 0 <= I5 - 1 /\ I5 <= 10]
1.59/1.69	  3) f2#(I6, I7, I8) -> f2#(0, 0, 9) [10 = I8 /\ I6 <= 1]
1.59/1.69	  4) f2#(I9, I10, I11) -> f2#(0, 0, I11 - 1) [0 = I10 /\ 0 <= I11 - 1 /\ I9 <= 1 /\ I11 <= 10 /\ I11 <= 9]
1.59/1.69	  5) f1#(I12, I13, I14) -> f2#(0, 0, I13) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
1.59/1.69	
1.59/1.69	We have the following SCCs.
1.59/1.69	  { 1, 2, 3, 4 }
1.59/1.69	
1.59/1.69	DP problem for innermost termination.
1.59/1.69	P =
1.59/1.69	  f2#(I0, I1, I2) -> f2#(1, 1, 1) [0 = I2 /\ I0 <= 1]
1.59/1.69	  f2#(I3, I4, I5) -> f2#(1, 1, I5 + 1) [1 = I4 /\ I3 <= 1 /\ I5 <= 9 /\ 0 <= I5 - 1 /\ I5 <= 10]
1.59/1.69	  f2#(I6, I7, I8) -> f2#(0, 0, 9) [10 = I8 /\ I6 <= 1]
1.59/1.69	  f2#(I9, I10, I11) -> f2#(0, 0, I11 - 1) [0 = I10 /\ 0 <= I11 - 1 /\ I9 <= 1 /\ I11 <= 10 /\ I11 <= 9]
1.59/1.69	R = 
1.59/1.69	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
1.59/1.69	  f2(I0, I1, I2) -> f2(1, 1, 1) [0 = I2 /\ I0 <= 1]
1.59/1.69	  f2(I3, I4, I5) -> f2(1, 1, I5 + 1) [1 = I4 /\ I3 <= 1 /\ I5 <= 9 /\ 0 <= I5 - 1 /\ I5 <= 10]
1.59/1.69	  f2(I6, I7, I8) -> f2(0, 0, 9) [10 = I8 /\ I6 <= 1]
1.59/1.69	  f2(I9, I10, I11) -> f2(0, 0, I11 - 1) [0 = I10 /\ 0 <= I11 - 1 /\ I9 <= 1 /\ I11 <= 10 /\ I11 <= 9]
1.59/1.69	  f1(I12, I13, I14) -> f2(0, 0, I13) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
1.59/1.69	
1.59/4.67	EOF