1.44/1.48	MAYBE
1.44/1.48	
1.44/1.48	DP problem for innermost termination.
1.44/1.48	P =
1.44/1.48	  init#(x1, x2) -> f1#(rnd1, rnd2) 
1.44/1.48	  f3#(I0, I1) -> f3#(I0 + 1, I1 - 2) [1 <= I1 - 1]
1.44/1.48	  f3#(I2, I3) -> f2#(I2, I4) [I3 <= 1]
1.44/1.48	  f2#(I5, I6) -> f3#(0, I5) [1 <= I5 - 1]
1.44/1.48	  f1#(I7, I8) -> f2#(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
1.44/1.48	R = 
1.44/1.48	  init(x1, x2) -> f1(rnd1, rnd2) 
1.44/1.48	  f3(I0, I1) -> f3(I0 + 1, I1 - 2) [1 <= I1 - 1]
1.44/1.48	  f3(I2, I3) -> f2(I2, I4) [I3 <= 1]
1.44/1.48	  f2(I5, I6) -> f3(0, I5) [1 <= I5 - 1]
1.44/1.48	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
1.44/1.48	
1.44/1.48	The dependency graph for this problem is:
1.44/1.48	  0 -> 4
1.44/1.48	  1 -> 1, 2
1.44/1.48	  2 -> 3
1.44/1.48	  3 -> 1
1.44/1.48	  4 -> 3
1.44/1.48	Where:
1.44/1.48	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
1.44/1.48	  1) f3#(I0, I1) -> f3#(I0 + 1, I1 - 2) [1 <= I1 - 1]
1.44/1.48	  2) f3#(I2, I3) -> f2#(I2, I4) [I3 <= 1]
1.44/1.48	  3) f2#(I5, I6) -> f3#(0, I5) [1 <= I5 - 1]
1.44/1.48	  4) f1#(I7, I8) -> f2#(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
1.44/1.48	
1.44/1.48	We have the following SCCs.
1.44/1.48	  { 1, 2, 3 }
1.44/1.48	
1.44/1.48	DP problem for innermost termination.
1.44/1.48	P =
1.44/1.48	  f3#(I0, I1) -> f3#(I0 + 1, I1 - 2) [1 <= I1 - 1]
1.44/1.48	  f3#(I2, I3) -> f2#(I2, I4) [I3 <= 1]
1.44/1.48	  f2#(I5, I6) -> f3#(0, I5) [1 <= I5 - 1]
1.44/1.48	R = 
1.44/1.48	  init(x1, x2) -> f1(rnd1, rnd2) 
1.44/1.48	  f3(I0, I1) -> f3(I0 + 1, I1 - 2) [1 <= I1 - 1]
1.44/1.48	  f3(I2, I3) -> f2(I2, I4) [I3 <= 1]
1.44/1.48	  f2(I5, I6) -> f3(0, I5) [1 <= I5 - 1]
1.44/1.48	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
1.44/1.48	
1.44/4.46	EOF