2.04/2.14	MAYBE
2.04/2.14	
2.04/2.14	DP problem for innermost termination.
2.04/2.14	P =
2.04/2.14	  f4#(x1, x2) -> f3#(x1, x2) 
2.04/2.14	  f3#(I0, I1) -> f1#(I0, I1) 
2.04/2.14	  f2#(I2, I3) -> f1#(-1 + I2, I3) 
2.04/2.14	  f2#(I4, I5) -> f1#(rnd1, -1 + I5) [rnd1 = rnd1]
2.04/2.14	  f1#(I6, I7) -> f2#(I6, I7) [1 <= I7 /\ 1 <= I6]
2.04/2.14	R = 
2.04/2.14	  f4(x1, x2) -> f3(x1, x2) 
2.04/2.14	  f3(I0, I1) -> f1(I0, I1) 
2.04/2.14	  f2(I2, I3) -> f1(-1 + I2, I3) 
2.04/2.14	  f2(I4, I5) -> f1(rnd1, -1 + I5) [rnd1 = rnd1]
2.04/2.14	  f1(I6, I7) -> f2(I6, I7) [1 <= I7 /\ 1 <= I6]
2.04/2.14	
2.04/2.14	The dependency graph for this problem is:
2.04/2.14	  0 -> 1
2.04/2.14	  1 -> 4
2.04/2.14	  2 -> 4
2.04/2.14	  3 -> 4
2.04/2.14	  4 -> 2, 3
2.04/2.14	Where:
2.04/2.14	  0) f4#(x1, x2) -> f3#(x1, x2) 
2.04/2.14	  1) f3#(I0, I1) -> f1#(I0, I1) 
2.04/2.14	  2) f2#(I2, I3) -> f1#(-1 + I2, I3) 
2.04/2.14	  3) f2#(I4, I5) -> f1#(rnd1, -1 + I5) [rnd1 = rnd1]
2.04/2.14	  4) f1#(I6, I7) -> f2#(I6, I7) [1 <= I7 /\ 1 <= I6]
2.04/2.14	
2.04/2.14	We have the following SCCs.
2.04/2.14	  { 2, 3, 4 }
2.04/2.14	
2.04/2.14	DP problem for innermost termination.
2.04/2.14	P =
2.04/2.14	  f2#(I2, I3) -> f1#(-1 + I2, I3) 
2.04/2.14	  f2#(I4, I5) -> f1#(rnd1, -1 + I5) [rnd1 = rnd1]
2.04/2.14	  f1#(I6, I7) -> f2#(I6, I7) [1 <= I7 /\ 1 <= I6]
2.04/2.14	R = 
2.04/2.14	  f4(x1, x2) -> f3(x1, x2) 
2.04/2.14	  f3(I0, I1) -> f1(I0, I1) 
2.04/2.14	  f2(I2, I3) -> f1(-1 + I2, I3) 
2.04/2.14	  f2(I4, I5) -> f1(rnd1, -1 + I5) [rnd1 = rnd1]
2.04/2.14	  f1(I6, I7) -> f2(I6, I7) [1 <= I7 /\ 1 <= I6]
2.04/2.14	
2.04/5.12	EOF