2.34/2.32	MAYBE
2.34/2.32	
2.34/2.32	DP problem for innermost termination.
2.34/2.32	P =
2.34/2.32	  f4#(x1, x2, x3) -> f3#(x1, x2, x3) 
2.34/2.32	  f3#(I0, I1, I2) -> f1#(rnd1, I1, I2) [y1 = -1 /\ y2 = -1 + y1 /\ y3 = -1 + y2 /\ y4 = -1 + y3 /\ y5 = -1 + y4 /\ y6 = -1 + y5 /\ y7 = -1 + y6 /\ y8 = -1 + y7 /\ y9 = -1 + y8 /\ rnd1 = -1 + y9 /\ -1 <= I1 /\ rnd1 <= I2]
2.34/2.32	  f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
2.34/2.32	  f1#(I6, I7, I8) -> f2#(I6, I7 + I8, I8) [I6 <= I8 /\ 0 <= I7]
2.34/2.32	R = 
2.34/2.32	  f4(x1, x2, x3) -> f3(x1, x2, x3) 
2.34/2.32	  f3(I0, I1, I2) -> f1(rnd1, I1, I2) [y1 = -1 /\ y2 = -1 + y1 /\ y3 = -1 + y2 /\ y4 = -1 + y3 /\ y5 = -1 + y4 /\ y6 = -1 + y5 /\ y7 = -1 + y6 /\ y8 = -1 + y7 /\ y9 = -1 + y8 /\ rnd1 = -1 + y9 /\ -1 <= I1 /\ rnd1 <= I2]
2.34/2.32	  f2(I3, I4, I5) -> f1(I3, I4, I5) 
2.34/2.32	  f1(I6, I7, I8) -> f2(I6, I7 + I8, I8) [I6 <= I8 /\ 0 <= I7]
2.34/2.32	
2.34/2.32	The dependency graph for this problem is:
2.34/2.32	  0 -> 1
2.34/2.32	  1 -> 3
2.34/2.32	  2 -> 3
2.34/2.32	  3 -> 2
2.34/2.32	Where:
2.34/2.32	  0) f4#(x1, x2, x3) -> f3#(x1, x2, x3) 
2.34/2.32	  1) f3#(I0, I1, I2) -> f1#(rnd1, I1, I2) [y1 = -1 /\ y2 = -1 + y1 /\ y3 = -1 + y2 /\ y4 = -1 + y3 /\ y5 = -1 + y4 /\ y6 = -1 + y5 /\ y7 = -1 + y6 /\ y8 = -1 + y7 /\ y9 = -1 + y8 /\ rnd1 = -1 + y9 /\ -1 <= I1 /\ rnd1 <= I2]
2.34/2.32	  2) f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
2.34/2.32	  3) f1#(I6, I7, I8) -> f2#(I6, I7 + I8, I8) [I6 <= I8 /\ 0 <= I7]
2.34/2.32	
2.34/2.32	We have the following SCCs.
2.34/2.32	  { 2, 3 }
2.34/2.32	
2.34/2.32	DP problem for innermost termination.
2.34/2.32	P =
2.34/2.32	  f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
2.34/2.32	  f1#(I6, I7, I8) -> f2#(I6, I7 + I8, I8) [I6 <= I8 /\ 0 <= I7]
2.34/2.32	R = 
2.34/2.32	  f4(x1, x2, x3) -> f3(x1, x2, x3) 
2.34/2.32	  f3(I0, I1, I2) -> f1(rnd1, I1, I2) [y1 = -1 /\ y2 = -1 + y1 /\ y3 = -1 + y2 /\ y4 = -1 + y3 /\ y5 = -1 + y4 /\ y6 = -1 + y5 /\ y7 = -1 + y6 /\ y8 = -1 + y7 /\ y9 = -1 + y8 /\ rnd1 = -1 + y9 /\ -1 <= I1 /\ rnd1 <= I2]
2.34/2.32	  f2(I3, I4, I5) -> f1(I3, I4, I5) 
2.34/2.32	  f1(I6, I7, I8) -> f2(I6, I7 + I8, I8) [I6 <= I8 /\ 0 <= I7]
2.34/2.32	
2.34/5.29	EOF