2.39/2.41	YES
2.39/2.41	
2.39/2.41	DP problem for innermost termination.
2.39/2.41	P =
2.39/2.41	  f5#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
2.39/2.41	  f4#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, 1000, I4) 
2.39/2.41	  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
2.39/2.41	  f1#(I10, I11, I12, I13, I14) -> f3#(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14]
2.39/2.41	R = 
2.39/2.41	  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
2.39/2.41	  f4(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1000, I4) 
2.39/2.41	  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
2.39/2.41	  f1(I10, I11, I12, I13, I14) -> f3(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14]
2.39/2.41	  f1(I15, I16, I17, I18, I19) -> f2(rnd1, rnd2, 0, I18, I19) [rnd1 = rnd2 /\ rnd2 = 0 /\ -100 + I18 <= 0 /\ 0 <= 9 - I19]
2.39/2.41	  f1(I20, I21, I22, I23, I24) -> f2(I25, I26, 0, I23, I24) [I25 = I26 /\ I26 = 0 /\ 10 - I24 <= 0]
2.39/2.41	
2.39/2.41	The dependency graph for this problem is:
2.39/2.41	  0 -> 1
2.39/2.41	  1 -> 3
2.39/2.41	  2 -> 3
2.39/2.41	  3 -> 2
2.39/2.41	Where:
2.39/2.41	  0) f5#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
2.39/2.41	  1) f4#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, 1000, I4) 
2.39/2.41	  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
2.39/2.41	  3) f1#(I10, I11, I12, I13, I14) -> f3#(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14]
2.39/2.41	
2.39/2.41	We have the following SCCs.
2.39/2.41	  { 2, 3 }
2.39/2.41	
2.39/2.41	DP problem for innermost termination.
2.39/2.41	P =
2.39/2.41	  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
2.39/2.41	  f1#(I10, I11, I12, I13, I14) -> f3#(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14]
2.39/2.41	R = 
2.39/2.41	  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
2.39/2.41	  f4(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1000, I4) 
2.39/2.41	  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
2.39/2.41	  f1(I10, I11, I12, I13, I14) -> f3(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14]
2.39/2.41	  f1(I15, I16, I17, I18, I19) -> f2(rnd1, rnd2, 0, I18, I19) [rnd1 = rnd2 /\ rnd2 = 0 /\ -100 + I18 <= 0 /\ 0 <= 9 - I19]
2.39/2.41	  f1(I20, I21, I22, I23, I24) -> f2(I25, I26, 0, I23, I24) [I25 = I26 /\ I26 = 0 /\ 10 - I24 <= 0]
2.39/2.41	
2.39/2.41	We use the basic value criterion with the projection function NU:
2.39/2.41	  NU[f1#(z1,z2,z3,z4,z5)] = z4
2.39/2.41	  NU[f3#(z1,z2,z3,z4,z5)] = z4
2.39/2.41	
2.39/2.41	This gives the following inequalities:
2.39/2.41	    ==>  I8 (>! \union =) I8
2.39/2.41	  0 <= -101 + I13 /\ 0 <= 9 - I14  ==>  I13 >! -1 + I13
2.39/2.41	
2.39/2.41	We remove all the strictly oriented dependency pairs.
2.39/2.41	
2.39/2.41	DP problem for innermost termination.
2.39/2.41	P =
2.39/2.41	  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
2.39/2.41	R = 
2.39/2.41	  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
2.39/2.41	  f4(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1000, I4) 
2.39/2.41	  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
2.39/2.41	  f1(I10, I11, I12, I13, I14) -> f3(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14]
2.39/2.41	  f1(I15, I16, I17, I18, I19) -> f2(rnd1, rnd2, 0, I18, I19) [rnd1 = rnd2 /\ rnd2 = 0 /\ -100 + I18 <= 0 /\ 0 <= 9 - I19]
2.39/2.41	  f1(I20, I21, I22, I23, I24) -> f2(I25, I26, 0, I23, I24) [I25 = I26 /\ I26 = 0 /\ 10 - I24 <= 0]
2.39/2.41	
2.39/2.41	The dependency graph for this problem is:
2.39/2.41	  2 -> 
2.39/2.41	Where:
2.39/2.41	  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
2.39/2.41	
2.39/2.41	We have the following SCCs.
2.39/2.41	  
2.39/5.39	EOF