1.09/1.42	MAYBE
1.09/1.42	
1.09/1.42	DP problem for innermost termination.
1.09/1.42	P =
1.09/1.42	  init#(x1, x2) -> f1#(rnd1, rnd2) 
1.09/1.42	  f2#(I0, I1) -> f2#(1, 1) [0 = I1]
1.09/1.42	  f2#(I2, I3) -> f2#(0, 9) [10 = I3]
1.09/1.42	  f2#(I4, I5) -> f2#(1, I5 + 1) [1 = I4 /\ 0 <= I5 - 1 /\ I5 <= 10 /\ I5 <= 9]
1.09/1.42	  f2#(I6, I7) -> f2#(0, I7 - 1) [0 = I6 /\ I7 <= 10 /\ 0 <= I7 - 1 /\ I7 <= 9]
1.09/1.42	  f1#(I8, I9) -> f2#(0, I9) [-1 <= I9 - 1 /\ 0 <= I8 - 1]
1.09/1.42	R = 
1.09/1.42	  init(x1, x2) -> f1(rnd1, rnd2) 
1.09/1.42	  f2(I0, I1) -> f2(1, 1) [0 = I1]
1.09/1.42	  f2(I2, I3) -> f2(0, 9) [10 = I3]
1.09/1.42	  f2(I4, I5) -> f2(1, I5 + 1) [1 = I4 /\ 0 <= I5 - 1 /\ I5 <= 10 /\ I5 <= 9]
1.09/1.42	  f2(I6, I7) -> f2(0, I7 - 1) [0 = I6 /\ I7 <= 10 /\ 0 <= I7 - 1 /\ I7 <= 9]
1.09/1.42	  f1(I8, I9) -> f2(0, I9) [-1 <= I9 - 1 /\ 0 <= I8 - 1]
1.09/1.42	
1.09/1.42	The dependency graph for this problem is:
1.09/1.42	  0 -> 5
1.09/1.42	  1 -> 3
1.09/1.42	  2 -> 4
1.09/1.42	  3 -> 2, 3
1.09/1.42	  4 -> 1, 4
1.09/1.42	  5 -> 1, 2, 4
1.09/1.42	Where:
1.09/1.42	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
1.09/1.42	  1) f2#(I0, I1) -> f2#(1, 1) [0 = I1]
1.09/1.42	  2) f2#(I2, I3) -> f2#(0, 9) [10 = I3]
1.09/1.42	  3) f2#(I4, I5) -> f2#(1, I5 + 1) [1 = I4 /\ 0 <= I5 - 1 /\ I5 <= 10 /\ I5 <= 9]
1.09/1.42	  4) f2#(I6, I7) -> f2#(0, I7 - 1) [0 = I6 /\ I7 <= 10 /\ 0 <= I7 - 1 /\ I7 <= 9]
1.09/1.42	  5) f1#(I8, I9) -> f2#(0, I9) [-1 <= I9 - 1 /\ 0 <= I8 - 1]
1.09/1.42	
1.09/1.42	We have the following SCCs.
1.09/1.42	  { 1, 2, 3, 4 }
1.09/1.42	
1.09/1.42	DP problem for innermost termination.
1.09/1.42	P =
1.09/1.42	  f2#(I0, I1) -> f2#(1, 1) [0 = I1]
1.09/1.42	  f2#(I2, I3) -> f2#(0, 9) [10 = I3]
1.09/1.42	  f2#(I4, I5) -> f2#(1, I5 + 1) [1 = I4 /\ 0 <= I5 - 1 /\ I5 <= 10 /\ I5 <= 9]
1.09/1.42	  f2#(I6, I7) -> f2#(0, I7 - 1) [0 = I6 /\ I7 <= 10 /\ 0 <= I7 - 1 /\ I7 <= 9]
1.09/1.42	R = 
1.09/1.42	  init(x1, x2) -> f1(rnd1, rnd2) 
1.09/1.42	  f2(I0, I1) -> f2(1, 1) [0 = I1]
1.09/1.42	  f2(I2, I3) -> f2(0, 9) [10 = I3]
1.09/1.42	  f2(I4, I5) -> f2(1, I5 + 1) [1 = I4 /\ 0 <= I5 - 1 /\ I5 <= 10 /\ I5 <= 9]
1.09/1.42	  f2(I6, I7) -> f2(0, I7 - 1) [0 = I6 /\ I7 <= 10 /\ 0 <= I7 - 1 /\ I7 <= 9]
1.09/1.42	  f1(I8, I9) -> f2(0, I9) [-1 <= I9 - 1 /\ 0 <= I8 - 1]
1.09/1.42	
1.09/4.07	EOF