0.75/0.77	MAYBE
0.75/0.77	
0.75/0.77	DP problem for innermost termination.
0.75/0.77	P =
0.75/0.77	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.75/0.77	  f2#(I0, I1) -> f2#(-1 * I0 - 1, I2) [-1 * I0 <= 1 /\ 4 <= I0 - 1]
0.75/0.77	  f2#(I3, I4) -> f2#(-1 * I3 + 1, I5) [I3 <= 4 /\ 0 <= I3 - 1]
0.75/0.77	  f2#(I6, I7) -> f2#(-1 * I6 + 1, I8) [I6 <= 4 /\ I6 <= -1]
0.75/0.77	  f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.75/0.77	R = 
0.75/0.77	  init(x1, x2) -> f1(rnd1, rnd2) 
0.75/0.77	  f2(I0, I1) -> f2(-1 * I0 - 1, I2) [-1 * I0 <= 1 /\ 4 <= I0 - 1]
0.75/0.77	  f2(I3, I4) -> f2(-1 * I3 + 1, I5) [I3 <= 4 /\ 0 <= I3 - 1]
0.75/0.77	  f2(I6, I7) -> f2(-1 * I6 + 1, I8) [I6 <= 4 /\ I6 <= -1]
0.75/0.77	  f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.75/0.77	
0.75/0.77	The dependency graph for this problem is:
0.75/0.77	  0 -> 4
0.75/0.77	  1 -> 3
0.75/0.77	  2 -> 3
0.75/0.77	  3 -> 1, 2
0.75/0.77	  4 -> 1, 2
0.75/0.77	Where:
0.75/0.77	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.75/0.77	  1) f2#(I0, I1) -> f2#(-1 * I0 - 1, I2) [-1 * I0 <= 1 /\ 4 <= I0 - 1]
0.75/0.77	  2) f2#(I3, I4) -> f2#(-1 * I3 + 1, I5) [I3 <= 4 /\ 0 <= I3 - 1]
0.75/0.77	  3) f2#(I6, I7) -> f2#(-1 * I6 + 1, I8) [I6 <= 4 /\ I6 <= -1]
0.75/0.77	  4) f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.75/0.77	
0.75/0.77	We have the following SCCs.
0.75/0.77	  { 1, 2, 3 }
0.75/0.77	
0.75/0.77	DP problem for innermost termination.
0.75/0.77	P =
0.75/0.77	  f2#(I0, I1) -> f2#(-1 * I0 - 1, I2) [-1 * I0 <= 1 /\ 4 <= I0 - 1]
0.75/0.77	  f2#(I3, I4) -> f2#(-1 * I3 + 1, I5) [I3 <= 4 /\ 0 <= I3 - 1]
0.75/0.77	  f2#(I6, I7) -> f2#(-1 * I6 + 1, I8) [I6 <= 4 /\ I6 <= -1]
0.75/0.77	R = 
0.75/0.77	  init(x1, x2) -> f1(rnd1, rnd2) 
0.75/0.77	  f2(I0, I1) -> f2(-1 * I0 - 1, I2) [-1 * I0 <= 1 /\ 4 <= I0 - 1]
0.75/0.77	  f2(I3, I4) -> f2(-1 * I3 + 1, I5) [I3 <= 4 /\ 0 <= I3 - 1]
0.75/0.77	  f2(I6, I7) -> f2(-1 * I6 + 1, I8) [I6 <= 4 /\ I6 <= -1]
0.75/0.77	  f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.75/0.77	
0.75/3.75	EOF