4.23/4.20	YES
4.23/4.20	
4.23/4.20	DP problem for innermost termination.
4.23/4.20	P =
4.23/4.20	  f7#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 
4.23/4.20	  f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
4.23/4.20	  f2#(I5, I6, I7, I8, I9) -> f6#(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
4.23/4.20	  f4#(I15, I16, I17, I18, I19) -> f2#(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10]
4.23/4.20	  f3#(I20, I21, I22, I23, I24) -> f2#(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1]
4.23/4.20	  f1#(I25, I26, I27, I28, I29) -> f2#(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0]
4.23/4.20	R = 
4.23/4.20	  f7(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
4.23/4.20	  f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
4.23/4.20	  f2(I5, I6, I7, I8, I9) -> f6(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
4.23/4.20	  f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I14, I14) [10 <= I11]
4.23/4.20	  f4(I15, I16, I17, I18, I19) -> f2(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10]
4.23/4.20	  f3(I20, I21, I22, I23, I24) -> f2(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1]
4.23/4.20	  f1(I25, I26, I27, I28, I29) -> f2(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0]
4.23/4.20	
4.23/4.20	The dependency graph for this problem is:
4.23/4.20	  0 -> 5
4.23/4.20	  1 -> 2
4.23/4.20	  2 -> 1
4.23/4.20	  3 -> 2
4.23/4.20	  4 -> 2
4.23/4.20	  5 -> 2
4.23/4.20	Where:
4.23/4.20	  0) f7#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 
4.23/4.20	  1) f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
4.23/4.20	  2) f2#(I5, I6, I7, I8, I9) -> f6#(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
4.23/4.20	  3) f4#(I15, I16, I17, I18, I19) -> f2#(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10]
4.23/4.20	  4) f3#(I20, I21, I22, I23, I24) -> f2#(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1]
4.23/4.20	  5) f1#(I25, I26, I27, I28, I29) -> f2#(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0]
4.23/4.20	
4.23/4.20	We have the following SCCs.
4.23/4.20	  { 1, 2 }
4.23/4.20	
4.23/4.20	DP problem for innermost termination.
4.23/4.20	P =
4.23/4.20	  f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
4.23/4.20	  f2#(I5, I6, I7, I8, I9) -> f6#(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
4.23/4.20	R = 
4.23/4.20	  f7(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
4.23/4.20	  f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
4.23/4.20	  f2(I5, I6, I7, I8, I9) -> f6(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
4.23/4.20	  f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I14, I14) [10 <= I11]
4.23/4.20	  f4(I15, I16, I17, I18, I19) -> f2(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10]
4.23/4.20	  f3(I20, I21, I22, I23, I24) -> f2(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1]
4.23/4.20	  f1(I25, I26, I27, I28, I29) -> f2(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0]
4.23/4.20	
4.23/4.20	We use the reverse value criterion with the projection function NU:
4.23/4.20	  NU[f2#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z2)
4.23/4.20	  NU[f6#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z2)
4.23/4.20	
4.23/4.20	This gives the following inequalities:
4.23/4.20	    ==>  10 + -1 * (1 + I1) >= 10 + -1 * (1 + I1)
4.23/4.20	  1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3  ==>  10 + -1 * (1 + I6) > 10 + -1 * (1 + (1 + I6))  with  10 + -1 * (1 + I6) >= 0
4.23/4.20	
4.23/4.20	We remove all the strictly oriented dependency pairs.
4.23/4.20	
4.23/4.20	DP problem for innermost termination.
4.23/4.20	P =
4.23/4.20	  f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
4.23/4.20	R = 
4.23/4.20	  f7(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
4.23/4.20	  f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
4.23/4.20	  f2(I5, I6, I7, I8, I9) -> f6(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
4.23/4.20	  f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I14, I14) [10 <= I11]
4.23/4.20	  f4(I15, I16, I17, I18, I19) -> f2(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10]
4.23/4.20	  f3(I20, I21, I22, I23, I24) -> f2(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1]
4.23/4.20	  f1(I25, I26, I27, I28, I29) -> f2(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0]
4.23/4.20	
4.23/4.20	The dependency graph for this problem is:
4.23/4.20	  1 -> 
4.23/4.20	Where:
4.23/4.20	  1) f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
4.23/4.20	
4.23/4.20	We have the following SCCs.
4.23/4.20	  
4.23/7.17	EOF