1.59/1.60	MAYBE
1.59/1.60	
1.59/1.60	DP problem for innermost termination.
1.59/1.60	P =
1.59/1.60	  f5#(x1, x2, x3) -> f1#(x1, x2, x3) 
1.59/1.60	  f3#(I3, I4, I5) -> f2#(I3, I4, I5) 
1.59/1.60	  f2#(I6, I7, I8) -> f3#(I6, I7 + I8, I8) [-1 * (I7 + I8) <= 0]
1.59/1.60	  f1#(I9, I10, I11) -> f2#(I9, I10, I11) 
1.59/1.60	R = 
1.59/1.60	  f5(x1, x2, x3) -> f1(x1, x2, x3) 
1.59/1.60	  f2(I0, I1, I2) -> f4(rnd1, I1 + I2, I2) [rnd1 = rnd1 /\ 0 <= -1 - (I1 + I2)]
1.59/1.60	  f3(I3, I4, I5) -> f2(I3, I4, I5) 
1.59/1.60	  f2(I6, I7, I8) -> f3(I6, I7 + I8, I8) [-1 * (I7 + I8) <= 0]
1.59/1.60	  f1(I9, I10, I11) -> f2(I9, I10, I11) 
1.59/1.60	
1.59/1.60	The dependency graph for this problem is:
1.59/1.60	  0 -> 3
1.59/1.60	  1 -> 2
1.59/1.60	  2 -> 1
1.59/1.60	  3 -> 2
1.59/1.60	Where:
1.59/1.60	  0) f5#(x1, x2, x3) -> f1#(x1, x2, x3) 
1.59/1.60	  1) f3#(I3, I4, I5) -> f2#(I3, I4, I5) 
1.59/1.60	  2) f2#(I6, I7, I8) -> f3#(I6, I7 + I8, I8) [-1 * (I7 + I8) <= 0]
1.59/1.60	  3) f1#(I9, I10, I11) -> f2#(I9, I10, I11) 
1.59/1.60	
1.59/1.60	We have the following SCCs.
1.59/1.60	  { 1, 2 }
1.59/1.60	
1.59/1.60	DP problem for innermost termination.
1.59/1.60	P =
1.59/1.60	  f3#(I3, I4, I5) -> f2#(I3, I4, I5) 
1.59/1.60	  f2#(I6, I7, I8) -> f3#(I6, I7 + I8, I8) [-1 * (I7 + I8) <= 0]
1.59/1.60	R = 
1.59/1.60	  f5(x1, x2, x3) -> f1(x1, x2, x3) 
1.59/1.60	  f2(I0, I1, I2) -> f4(rnd1, I1 + I2, I2) [rnd1 = rnd1 /\ 0 <= -1 - (I1 + I2)]
1.59/1.60	  f3(I3, I4, I5) -> f2(I3, I4, I5) 
1.59/1.60	  f2(I6, I7, I8) -> f3(I6, I7 + I8, I8) [-1 * (I7 + I8) <= 0]
1.59/1.60	  f1(I9, I10, I11) -> f2(I9, I10, I11) 
1.59/1.60	
1.59/4.58	EOF