4.18/4.14	MAYBE
4.18/4.14	
4.18/4.14	DP problem for innermost termination.
4.18/4.14	P =
4.18/4.14	  f6#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 
4.18/4.14	  f5#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
4.18/4.14	  f2#(I5, I6, I7, I8, I9) -> f5#(I5, -1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
4.18/4.14	  f3#(I15, I16, I17, I18, I19) -> f2#(rnd1, rnd2, I17, I18, I19) [1 + I16 <= 10 /\ y1 = -1 + I16 /\ rnd2 = rnd2 /\ -1 <= rnd2 /\ rnd2 <= -1 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1]
4.18/4.14	  f1#(I20, I21, I22, I23, I24) -> f2#(I20, 0, I22, I23, I24) [0 <= 0 /\ 0 <= 0]
4.18/4.14	R = 
4.18/4.14	  f6(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
4.18/4.14	  f5(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
4.18/4.14	  f2(I5, I6, I7, I8, I9) -> f5(I5, -1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
4.18/4.14	  f2(I10, I11, I12, I13, I14) -> f4(I10, I11, I12, I14, I14) [10 <= I11]
4.18/4.14	  f3(I15, I16, I17, I18, I19) -> f2(rnd1, rnd2, I17, I18, I19) [1 + I16 <= 10 /\ y1 = -1 + I16 /\ rnd2 = rnd2 /\ -1 <= rnd2 /\ rnd2 <= -1 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1]
4.18/4.14	  f1(I20, I21, I22, I23, I24) -> f2(I20, 0, I22, I23, I24) [0 <= 0 /\ 0 <= 0]
4.18/4.14	
4.18/4.14	The dependency graph for this problem is:
4.18/4.14	  0 -> 4
4.18/4.14	  1 -> 2
4.18/4.14	  2 -> 1
4.18/4.14	  3 -> 2
4.18/4.14	  4 -> 2
4.18/4.14	Where:
4.18/4.14	  0) f6#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 
4.18/4.14	  1) f5#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
4.18/4.14	  2) f2#(I5, I6, I7, I8, I9) -> f5#(I5, -1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
4.18/4.14	  3) f3#(I15, I16, I17, I18, I19) -> f2#(rnd1, rnd2, I17, I18, I19) [1 + I16 <= 10 /\ y1 = -1 + I16 /\ rnd2 = rnd2 /\ -1 <= rnd2 /\ rnd2 <= -1 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1]
4.18/4.14	  4) f1#(I20, I21, I22, I23, I24) -> f2#(I20, 0, I22, I23, I24) [0 <= 0 /\ 0 <= 0]
4.18/4.14	
4.18/4.14	We have the following SCCs.
4.18/4.14	  { 1, 2 }
4.18/4.14	
4.18/4.14	DP problem for innermost termination.
4.18/4.14	P =
4.18/4.14	  f5#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
4.18/4.14	  f2#(I5, I6, I7, I8, I9) -> f5#(I5, -1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
4.18/4.14	R = 
4.18/4.14	  f6(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
4.18/4.14	  f5(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
4.18/4.14	  f2(I5, I6, I7, I8, I9) -> f5(I5, -1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
4.18/4.14	  f2(I10, I11, I12, I13, I14) -> f4(I10, I11, I12, I14, I14) [10 <= I11]
4.18/4.14	  f3(I15, I16, I17, I18, I19) -> f2(rnd1, rnd2, I17, I18, I19) [1 + I16 <= 10 /\ y1 = -1 + I16 /\ rnd2 = rnd2 /\ -1 <= rnd2 /\ rnd2 <= -1 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1]
4.18/4.14	  f1(I20, I21, I22, I23, I24) -> f2(I20, 0, I22, I23, I24) [0 <= 0 /\ 0 <= 0]
4.18/4.14	
4.18/7.12	EOF