1.08/1.14	MAYBE
1.08/1.14	
1.08/1.14	DP problem for innermost termination.
1.08/1.14	P =
1.08/1.14	  init#(x1, x2) -> f1#(rnd1, rnd2) 
1.08/1.14	  f3#(I0, I1) -> f2#(I2, I3) [I0 <= 14 /\ I0 + 1 - 10 * y1 <= 9 /\ 0 <= I0 + 1 - 10 * y1 /\ I0 + 1 - 10 * y1 = I2]
1.08/1.14	  f2#(I4, I5) -> f3#(I4, I6) [I4 <= 14]
1.08/1.14	  f1#(I7, I8) -> f2#(I8, I9) [-1 <= I8 - 1 /\ 0 <= I7 - 1]
1.08/1.14	R = 
1.08/1.14	  init(x1, x2) -> f1(rnd1, rnd2) 
1.08/1.14	  f3(I0, I1) -> f2(I2, I3) [I0 <= 14 /\ I0 + 1 - 10 * y1 <= 9 /\ 0 <= I0 + 1 - 10 * y1 /\ I0 + 1 - 10 * y1 = I2]
1.08/1.14	  f2(I4, I5) -> f3(I4, I6) [I4 <= 14]
1.08/1.14	  f1(I7, I8) -> f2(I8, I9) [-1 <= I8 - 1 /\ 0 <= I7 - 1]
1.08/1.14	
1.08/1.14	The dependency graph for this problem is:
1.08/1.14	  0 -> 3
1.08/1.14	  1 -> 2
1.08/1.14	  2 -> 1
1.08/1.14	  3 -> 2
1.08/1.14	Where:
1.08/1.14	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
1.08/1.14	  1) f3#(I0, I1) -> f2#(I2, I3) [I0 <= 14 /\ I0 + 1 - 10 * y1 <= 9 /\ 0 <= I0 + 1 - 10 * y1 /\ I0 + 1 - 10 * y1 = I2]
1.08/1.14	  2) f2#(I4, I5) -> f3#(I4, I6) [I4 <= 14]
1.08/1.14	  3) f1#(I7, I8) -> f2#(I8, I9) [-1 <= I8 - 1 /\ 0 <= I7 - 1]
1.08/1.14	
1.08/1.14	We have the following SCCs.
1.08/1.14	  { 1, 2 }
1.08/1.14	
1.08/1.14	DP problem for innermost termination.
1.08/1.14	P =
1.08/1.14	  f3#(I0, I1) -> f2#(I2, I3) [I0 <= 14 /\ I0 + 1 - 10 * y1 <= 9 /\ 0 <= I0 + 1 - 10 * y1 /\ I0 + 1 - 10 * y1 = I2]
1.08/1.14	  f2#(I4, I5) -> f3#(I4, I6) [I4 <= 14]
1.08/1.14	R = 
1.08/1.14	  init(x1, x2) -> f1(rnd1, rnd2) 
1.08/1.14	  f3(I0, I1) -> f2(I2, I3) [I0 <= 14 /\ I0 + 1 - 10 * y1 <= 9 /\ 0 <= I0 + 1 - 10 * y1 /\ I0 + 1 - 10 * y1 = I2]
1.08/1.14	  f2(I4, I5) -> f3(I4, I6) [I4 <= 14]
1.08/1.14	  f1(I7, I8) -> f2(I8, I9) [-1 <= I8 - 1 /\ 0 <= I7 - 1]
1.08/1.14	
1.08/1.14	EOF