0.54/0.65	MAYBE
0.54/0.65	
0.54/0.65	DP problem for innermost termination.
0.54/0.65	P =
0.54/0.65	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.54/0.65	  f2#(I0, I1) -> f2#(-1 * I0 + 2, I2) [2 <= I0 - 1]
0.54/0.65	  f2#(I3, I4) -> f2#(-1 * I3 - 2, I5) [I3 <= -1 /\ I3 <= -1 - 1 /\ I3 <= -2 - 1]
0.54/0.65	  f2#(I6, I7) -> f2#(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1]
0.54/0.65	  f2#(I9, I10) -> f2#(I9 + 2, I11) [-3 <= I9 - 1 /\ I9 <= -1]
0.54/0.65	  f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
0.54/0.65	R = 
0.54/0.65	  init(x1, x2) -> f1(rnd1, rnd2) 
0.54/0.65	  f2(I0, I1) -> f2(-1 * I0 + 2, I2) [2 <= I0 - 1]
0.54/0.65	  f2(I3, I4) -> f2(-1 * I3 - 2, I5) [I3 <= -1 /\ I3 <= -1 - 1 /\ I3 <= -2 - 1]
0.54/0.65	  f2(I6, I7) -> f2(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1]
0.54/0.65	  f2(I9, I10) -> f2(I9 + 2, I11) [-3 <= I9 - 1 /\ I9 <= -1]
0.54/0.65	  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
0.54/0.65	
0.54/0.65	The dependency graph for this problem is:
0.54/0.65	  0 -> 5
0.54/0.65	  1 -> 2, 4
0.54/0.65	  2 -> 1, 3
0.54/0.65	  3 -> 4
0.54/0.65	  4 -> 3
0.54/0.65	  5 -> 1, 3
0.54/0.65	Where:
0.54/0.65	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.54/0.65	  1) f2#(I0, I1) -> f2#(-1 * I0 + 2, I2) [2 <= I0 - 1]
0.54/0.65	  2) f2#(I3, I4) -> f2#(-1 * I3 - 2, I5) [I3 <= -1 /\ I3 <= -1 - 1 /\ I3 <= -2 - 1]
0.64/0.65	  3) f2#(I6, I7) -> f2#(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1]
0.64/0.65	  4) f2#(I9, I10) -> f2#(I9 + 2, I11) [-3 <= I9 - 1 /\ I9 <= -1]
0.64/0.65	  5) f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
0.64/0.65	
0.64/0.65	We have the following SCCs.
0.64/0.65	  { 1, 2 }
0.64/0.65	  { 3, 4 }
0.64/0.65	
0.64/0.65	DP problem for innermost termination.
0.64/0.65	P =
0.64/0.65	  f2#(I6, I7) -> f2#(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1]
0.64/0.65	  f2#(I9, I10) -> f2#(I9 + 2, I11) [-3 <= I9 - 1 /\ I9 <= -1]
0.64/0.65	R = 
0.64/0.65	  init(x1, x2) -> f1(rnd1, rnd2) 
0.64/0.65	  f2(I0, I1) -> f2(-1 * I0 + 2, I2) [2 <= I0 - 1]
0.64/0.65	  f2(I3, I4) -> f2(-1 * I3 - 2, I5) [I3 <= -1 /\ I3 <= -1 - 1 /\ I3 <= -2 - 1]
0.64/0.65	  f2(I6, I7) -> f2(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1]
0.64/0.65	  f2(I9, I10) -> f2(I9 + 2, I11) [-3 <= I9 - 1 /\ I9 <= -1]
0.64/0.65	  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
0.64/0.65	
0.64/3.64	EOF