2.31/2.35	YES
2.31/2.35	
2.31/2.35	DP problem for innermost termination.
2.31/2.35	P =
2.31/2.35	  f5#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 
2.31/2.35	  f4#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
2.31/2.35	  f2#(I5, I6, I7, I8, I9) -> f4#(I5, I6, I7, -1 + I8, I9) [1 <= I9 /\ -1 + I9 <= -1 + I8 /\ -1 + I8 <= -1 + I9 /\ 1 <= I8]
2.31/2.35	  f1#(I15, I16, I17, I18, I19) -> f2#(rnd1, I16, I17, rnd4, I19) [y1 = y1 /\ rnd4 = y1 /\ rnd1 = rnd1]
2.31/2.35	R = 
2.31/2.35	  f5(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
2.31/2.35	  f4(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
2.31/2.35	  f2(I5, I6, I7, I8, I9) -> f4(I5, I6, I7, -1 + I8, I9) [1 <= I9 /\ -1 + I9 <= -1 + I8 /\ -1 + I8 <= -1 + I9 /\ 1 <= I8]
2.31/2.35	  f2(I10, I11, I12, I13, I14) -> f3(I10, I12, I12, I13, I14) [I13 <= 0]
2.31/2.35	  f1(I15, I16, I17, I18, I19) -> f2(rnd1, I16, I17, rnd4, I19) [y1 = y1 /\ rnd4 = y1 /\ rnd1 = rnd1]
2.31/2.35	
2.31/2.35	The dependency graph for this problem is:
2.31/2.35	  0 -> 3
2.31/2.35	  1 -> 2
2.31/2.35	  2 -> 1
2.31/2.35	  3 -> 2
2.31/2.35	Where:
2.31/2.35	  0) f5#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 
2.31/2.35	  1) f4#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
2.31/2.35	  2) f2#(I5, I6, I7, I8, I9) -> f4#(I5, I6, I7, -1 + I8, I9) [1 <= I9 /\ -1 + I9 <= -1 + I8 /\ -1 + I8 <= -1 + I9 /\ 1 <= I8]
2.31/2.35	  3) f1#(I15, I16, I17, I18, I19) -> f2#(rnd1, I16, I17, rnd4, I19) [y1 = y1 /\ rnd4 = y1 /\ rnd1 = rnd1]
2.31/2.35	
2.31/2.35	We have the following SCCs.
2.31/2.35	  { 1, 2 }
2.31/2.35	
2.31/2.35	DP problem for innermost termination.
2.31/2.35	P =
2.31/2.35	  f4#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
2.31/2.35	  f2#(I5, I6, I7, I8, I9) -> f4#(I5, I6, I7, -1 + I8, I9) [1 <= I9 /\ -1 + I9 <= -1 + I8 /\ -1 + I8 <= -1 + I9 /\ 1 <= I8]
2.31/2.35	R = 
2.31/2.35	  f5(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
2.31/2.35	  f4(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
2.31/2.35	  f2(I5, I6, I7, I8, I9) -> f4(I5, I6, I7, -1 + I8, I9) [1 <= I9 /\ -1 + I9 <= -1 + I8 /\ -1 + I8 <= -1 + I9 /\ 1 <= I8]
2.31/2.35	  f2(I10, I11, I12, I13, I14) -> f3(I10, I12, I12, I13, I14) [I13 <= 0]
2.31/2.35	  f1(I15, I16, I17, I18, I19) -> f2(rnd1, I16, I17, rnd4, I19) [y1 = y1 /\ rnd4 = y1 /\ rnd1 = rnd1]
2.31/2.35	
2.31/2.35	We use the basic value criterion with the projection function NU:
2.31/2.35	  NU[f2#(z1,z2,z3,z4,z5)] = z4
2.31/2.35	  NU[f4#(z1,z2,z3,z4,z5)] = z4
2.31/2.35	
2.31/2.35	This gives the following inequalities:
2.31/2.35	    ==>  I3 (>! \union =) I3
2.31/2.35	  1 <= I9 /\ -1 + I9 <= -1 + I8 /\ -1 + I8 <= -1 + I9 /\ 1 <= I8  ==>  I8 >! -1 + I8
2.31/2.35	
2.31/2.35	We remove all the strictly oriented dependency pairs.
2.31/2.35	
2.31/2.35	DP problem for innermost termination.
2.31/2.35	P =
2.31/2.35	  f4#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
2.31/2.35	R = 
2.31/2.35	  f5(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
2.31/2.35	  f4(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
2.31/2.35	  f2(I5, I6, I7, I8, I9) -> f4(I5, I6, I7, -1 + I8, I9) [1 <= I9 /\ -1 + I9 <= -1 + I8 /\ -1 + I8 <= -1 + I9 /\ 1 <= I8]
2.31/2.35	  f2(I10, I11, I12, I13, I14) -> f3(I10, I12, I12, I13, I14) [I13 <= 0]
2.31/2.35	  f1(I15, I16, I17, I18, I19) -> f2(rnd1, I16, I17, rnd4, I19) [y1 = y1 /\ rnd4 = y1 /\ rnd1 = rnd1]
2.31/2.35	
2.31/2.35	The dependency graph for this problem is:
2.31/2.35	  1 -> 
2.31/2.35	Where:
2.31/2.35	  1) f4#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
2.31/2.35	
2.31/2.35	We have the following SCCs.
2.31/2.35	  
2.31/5.33	EOF